# Linear Algebra II: Eigenvectors and Diagonalisability

This post continues the discussion of the Oxford first-year course Linear Algebra II. We’ve moved on from determinants, and are now considering eigenvalues and eigenvectors of matrices and linear maps.

A good question to ask is: what’s the point of knowing about eigenvectors? I can think of a quick answer and a longer answer. The quick answer is that whenever we have a mapping of any kind, it is natural to ask about its fixed points. And since we are thinking about vector spaces and linear maps, if we can’t find any fixed points, we might nonetheless be able to find the best thing, some vectors whose direction is fixed by the map. In general, knowing about fixed points of a mapping might tell us other more qualitative properties, including the behaviour seen when you apply the map iteratively a large number of times. (Indeed a recent post discusses this exact problem for positive matrices in a context relevant to a chapter of my thesis…)

A more specific answer concerns bases. Recall that a linear map is defined independently of any basis: it’s just a map from the vector space to itself. We can express the linear map via a matrix with respect to some basis, but how to choose the basis? We could always choose the canonical basis in $\mathbb{R}^n$, since it’s easy to do vector and matrix calculations when most of the entries of all the vectors are zero. We also have a good visual idea (at least in up to three dimensions) of what a matrix might mean with respect to that basis. If we needed to divide the three-dimensional world around us into small volumes, we’d tend to describe it with small cubes rather than small arbitrary parallelopipeds.

But once we know something about the linear map, we might want to choose a basis of vectors on which the behaviour of the map is particularly easy to describe. And eigenvectors fulfil precisely this role. If we are able to choose a basis of eigenvectors, describing the map’s action, either abstractly, or via a (diagonal) matrix, is very straightforward. If we are given a matrix to begin with, we know how to do a change of basis, and changing to the basis of eigenvectors is precisely what’s going when we write $A=P^{-1}DP$, where D is a diagonal matrix. We construct P by taking its columns to be these eigenvectors. In particular, for a given vector x, y=Px is the vector giving the coefficients of x in the basis of eigenvectors.

So the case where we have a basis of eigenvectors is particularly useful, and in this case, we say the matrix or the map is diagonalisable. Remember how we find eigenvalues. If there exists a non-zero vector x satisfying $Ax=\lambda x$, then x is in the kernel of $A-\lambda I$. As we discussed last time, introducing the determinant gives a much more manageable way to verify which values of $\lambda$ result in $A-\lambda I$ having a non-trivial kernel. In particular, if non-zero x is in the kernel, we have $\mathrm{det}(A-\lambda I)=0$, and this leads to a polynomial of degree n (the dimensional of the vector space / size of the matrix) for $\lambda$, called the characteristic polynomial $\chi_A(z)$, which has the eigenvalues as its roots.

If we agree to work over the complex field, then this is good, because it means we always have eigenvalues, and so it becomes sensible to talk about exactly how many eigenvalues and eigenvectors we have. Observe that if we restrict to real vector spaces, this might not be the case. In the plane, the rotation by $\pi/2$ for example has no fixed vectors.

Multiplicities of eigenvalues

We call the algebraic multiplicity $\alpha(\lambda)$ of an eigenvalue $\lambda$ to be the exponent of the factor $(z-\lambda)$ in the factorisation of the characteristic polynomial. To define the geometric multiplicity, observe that all the eigenvectors with eigenvalue $\lambda$ form a subspace, and so it is meaningful to talk about the dimension of this subspace (‘eigenspace’), which is the geometric multiplicity $\gamma(\lambda)$. There are two facts that one needs to remember. The slightly less obvious one is that $\gamma(\lambda)\le \alpha(\lambda)$ for all $\lambda$. One can see this by, for example, working in a basis that extends a basis of the $\lambda$-eigenspace. Observe at this stage that the sum of the algebraic multiplicities has to be n by definition, while the sum of geometric multiplicities is at most n. And this makes sense, because the space spanned by all the eigenvectors is a subspace, and so has dimension at most n.

The more obvious, but more frequently forgotten result is that

$\alpha(\lambda)\ge 1 \quad \iff \quad \gamma(\lambda)\ge 1,$

which is simply a consequence of the property discussed a few paragraphs previously concerning the kernel of $A-\lambda I$.

In particular, we might make the heuristic observation that ‘most’ polynomials of degree n have n distinct roots. This is certainly true for quadratics: there is only one value that the discriminant can take such that we see a repeated root. Alternatively, imagine shifting the quadratic up and down (in a complex way if necessary); again there is only one moment at which it might have a repeated root. This observation can be generalised easily to higher degree polynomials in a number of ways.

So if we lift this observation across to matrices, we see that most matrices have n distinct eigenvalues, and thus have n linearly independent eigenvectors which form a basis, hence the matrix is diagonalisable. I think it’s really worth reflecting on this, since much of a first exploration into linear algebra ends up treating exactly the case where the matrix is not diagonalisable.

The principal example of a non-diagonalisable matrix is $\begin{pmatrix}2&1\\0&2\end{pmatrix}$, where the 2s can be replaced by any value, and the 1 can be replaced by an non-zero value. There’s plenty to learn about to what extent versions of this matrix of higher size represent all non-diagonalisable matrices, but such an exposition of Jordan normal form comes next year for the students taking this course.

It probably is worth saying now though, that this example gives a good sanity check for whether a method is actually using diagonalisability correctly. For example, it is easily seen that elementary row operations to not preserve diagonalisability by starting from $\begin{pmatrix}2&0\\0&2\end{pmatrix}$ and ending up at our counter-example. One could also argue from this that the set of non-diagonalisable matrices are dense within the set of matrices with a repeated eigenvalue. That is, having a repeated eigenvalue but full eigenspace is doubly-infinitely-unlikely.

Cayley-Hamilton theorem

Anyway, among other results, we also saw the Cayley-Hamilton theorem, which states that a matrix A satisfies its own characteristic equation. That is $\chi_A(A)=0$, where the zero on the right-hand side is the zero matrix. It’s tempting to substitute A into the expression $\mathrm{det}(A-\lambda I)$, but of course this is not valid. Indeed imagine a typical eigenvalue determinant matrix with terms like $(7-\lambda)$ on the diagonal; it doesn’t make sense to substitute a matrix for $\lambda$ as one of the entries of the overall matrix!

Fortunately, we can argue convincingly in the case where A is a diagonalisable matrix. Remember that $\chi_A(A)$ is a matrix. Now looki at the action of $\chi_A(A)$ on any eigenvector v, corresponding to eigenvalue $\lambda$. Applying some power of A to v gives v multiplied by the same power of $\lambda$, and so we end up with

$\chi_A(A)v = \chi_A(\lambda)v = 0.$

This only worked when v was an eigenvector, but fortunately there is a basis of eigenvectors if A is diagonalisable, and so $\chi_A(A)v=0$ for all v, hence $\chi_A(A)=0$.

But $\chi_A(A)$ is just a matrix-valued function of A. If you think about it, $\chi_A$ is a monic polynomial, all of whose non-leading coefficients are multinomials of degree at most n-1 in the entries of A. Furthermore, these multinomials have (non-negative) integer coefficients. Therefore the entries of $\chi_A(A)$ are multinomials of degree at most 2n-1 in the entries of A, and again have (non-negative) integer coefficients.

Even without the integrality of the coefficients, this says that, under any reasonable definition of continuity of matrices (which could be induced from any topology on $\mathbb{R}^{n\times n}$) the function $\chi_A(A)$ should be continuous as a function of A. But we’ve shown $\chi_A(A)=0$ for all diagonalisable A, and also argued that most complex-valued matrices are diagonalisable. Turning this into a formal statement about denseness means that we’ve shown the Cayley-Hamilton theorem for non-diagonalisable matrices also. It feels that because the coefficients are non-negative integers, we might also have shown the result for other fields too, but I have minimal knowledge or recollection at the moment of the things one has to check for this sort of result.

It’s worth ending with the brief comment that Cayley-Hamilton is useful, among other reasons because it enables us to write the inverse of A as a polynomial of degree at most n-1 in terms of A. In many settings this is a lot easier to work with in terms of calculations than an argument with minors.

# Linear Algebra II: Determinants 2

In the previous post, we introduced determinants of matrices (and by extension linear maps) via its multilinearity properties, and as the change-of-volume factor. We also discussed how to calculate them, via row operations, or Laplace expansion, or directly via a sum of products of entries over permutations.

The question of why this is ever a useful quantity to consider remains, and this post tries to answer it. We’ll start by seeing one example where this is a very natural quantity to consider, and then the main abstract setting, where the determinant is zero, and consider a particularly nice example of this.

Jacobeans as a determinant

We consider integration by substitution. Firstly, in one variable: when it comes to Riemann integration of a function g(x) with respect to x, we view dx as the width of a small column which approximates the function near x. Now, if we reparameterise, that is if we write x=f(y) for some well-behaved (in particular differentiable) function f, then the width of the column is dx= dy.(dx/dy)=f'(y) dy. This may be negative, if y is decreasing while x is increasing, but for now let’s not worry about this overly, for example by assuming the function g is non-negative. Thus if we want to integrate with y as the variable, we multiply the integrand by this factor $|f'(y)|$.

What about in higher dimensions? We have exactly the same situation, only instead of two-dimensional columns, we have (n+1)-dimensional columns. We then multiply the n-dimensional volume of the base by the height, again given by $g(\mathbf{x})$. If we have a similar transformation of the base variable $\mathbf{x}=f(\mathbf{y})$, we differentiate to get

$\mathrm{d}x_i = \sum_{j=1}^n\frac{\mathrm{d}f_i}{\mathrm{d}y_j} \mathrm{d}y_j.$

In other words

$\mathrm{d}\mathbf{x}= J \mathrm{d}\mathbf{y},$

where J is the Jacobean matrix of partial deriatives. In particular, we know how to relate the volume $[0,\mathrm{d}x_1]\times\ldots\times [0,\mathrm{d}x_n]$ to the volume $[0,\mathrm{d}y_1]\times \ldots\times [0,\mathrm{d}y_n]$. It’s simply the determinant of the Jacobean J. So if we want to integrate with respect to $\mathbf{y}$, it only remains to pre-multiply the integrand by $|\mathrm{det}J|$ and proceed otherwise as in the one-dimensional case.

Det A = 0

A first linear algebra course might well motivate the introducing matrices as a notational shortcut for solving families of linear equations, $Ax=b$. The main idea is that generally we can solve this equation uniquely. Almost all of the theory developed in such a first linear algebra course deals with the case when this fails to hold. In particular, there are many ways to characterise this case, and we list some of them now:

• Ax=b has no solutions for some b;
• A is not invertible;
• A has non-trivial kernel, that is, with dimension at least one;
• A does not have full rank, that is, the image has dimension less than n;
• The columns (or indeed the rows) are linearly dependent;
• The matrix can be row-reduced to a matrix with a row of zeroes.

It is useful that these are equivalent, as in abstract problems one can choose whichever interpretation from this list is most relevant. However, all of these are quite hard to check. Exhibiting a non-trivial kernel element is hard – one either has to do manual row-reduction, or the equivalent in the context of linear equations. But we can add the characterisation

• det A = 0;

to the list. And this is genuinely much easier to check for specific examples, either abstract or numerical.

Let’s quickly convince ourselves of a couple of these equivalences. Determinant is invariant under row-reductions, and by multilinearity it is certainly the case that det A = 0 if A has a row of zeroes. We also said that A is the change-of-volume factor. Note that A is a map from the domain to its image, so if A has less than full rank, then any set in the image has zero volume.

The Vandermonde matrix

This is a good example of this theory in practice. Consider the Vandermonde matrix where each row is a geometric progression:

$V=\begin{pmatrix}1&\alpha_1&\ldots&\alpha_1^{n-1}\\1&\alpha_2&\ldots&\alpha_2^{n-1}\\ \vdots&\vdots&\ddots&\vdots\\ 1&\alpha_n&\ldots&\alpha_n^{n-1}\end{pmatrix}.$

Now suppose we attempt to solve

$V\begin{pmatrix}a_0\\a_1\\ \vdots\\ a_{n-1}\end{pmatrix}=\begin{pmatrix}b_1\\b_2\\ \vdots \\ b_n\end{pmatrix}.$

There’s a natural interpretation to this, that’s especially clear with this suggestive notation. Each row corresponds to a polynomial, where the coefficients are given by the $(a_0,a_1,\ldots,a_{n-1})$, and the argument is given by $\alpha_i$.

So if we try to solve for $(a_0,a_1,\ldots,a_{n-1})$, given $(\alpha_1,\ldots,\alpha_n)$ and $(b_1,\ldots,b_n)$, we are asking whether we can find a polynomial P with degree at most n-1 such that $P(\alpha_i)=b_i$ for \$i=1. Lagrange interpolation gives an argument where we just directly write down the relevant polynomial, but we can also deploy our linear algebraic arguments too.

The equivalence of all these statements means that to verify existence and uniqueness of such a polynomial, we only need to check that the Vandermonde matrix has non-zero determinant. And in fact there are a variety of methods to show that

$\mathrm{det}V=\prod_{1\le i< j}\le n(\alpha_j-\alpha_i).$

For the polynomial question to be meaningful, we would certainly demand that the $(\alpha_i)$ are distinct, and so this determinant is non-zero, and we’ve shown that n points determine a degree (n-1) polynomial uniquely.

If we multiply on the left instead, suppose that we are considering a discrete probability distribution X that takes n known values $(\alpha_1,\ldots,\alpha_n)$ with unknown probabilities $(p_1,\ldots,p_n)$. Then we have

$(p_1,\ldots,p_n) V = (1,\mathbb{E}X, \mathbb{E}[X^2],\ldots, \mathbb{E}[X^{n-1}]).$

So, again by inverting the Vandermonde matrix (which is know is possible since its determinant is non-zero…) we can recover the distribution from the first (n-1) moments of the distribution.

A similar argument applies to show that the Discrete Fourier Transform is invertible, and in this case (where the $\alpha_i$s are roots of unity), the expression for the Vandermonde determinant is particularly tractable.

# Linear Algebra II: Determinants 1

This term, I’m giving tutorials on a course that’s new to me, the apparently notorious ‘Linear Algebra II’ for first year undergraduates. I can appreciate how it might have ended up with this reputation, but as always, every challenge is also an opportunity, So I’m going to (try to) write a short series of blog posts about what we’ve discussed in the tutorials.

The first problem sheet-and-a-half concerned determinants of matrices. There are three things worth addressing here:

1. What are abstract definitions, and which is most useful in each setting?
2. How to actually calculate them?
3. What’s the overall point?

The answers are obviously not completely unrelated, but we’ll probably defer the third question to a second post.

The determinant is a map from the set of matrices $\mathcal{M}_n$ to the base field (hereafter assumed to be $\mathbb{R},\mathbb{C}$). The Oxford course defines it through its properties:

• Multilinear in the columns of the matrix.
• Equal to zero if two columns are equal.
• Equal to one if the matrix is the identity.

One then checks that there is a unique such map, and so from now on it’s reasonable to call it the determinant of the matrix. It will follow from pretty much any consequence that we can replace ‘columns’ with ‘rows’ throughout and get the same map.

Other definitions

We have a closed form expression for the determinant given via permutations of n

$\mathrm{det}(A)=\sum_{\sigma\in \Sigma_n} \mathrm{sign}(\sigma) a_{1\sigma(1)}\ldots a_{n\sigma(n)}.$

We’ll come back to a discussion of when this particular definition is useful. It can be derived by carefully transforming the identity matrix into A, using the operations which are mentioned in the original definition of the determinant, in particular, keeping track of the number of transpositions of columns.

It’s clear from any definition that the determinant is a polynomial of degree n in the entries of the matrix, but this definition will be useful if you want to make some more precise comment on the nature of this polynomial. For example, if entries of the matrix are polynomials in x of various degree (think of the eigenvalue equation for example) this allows you to control (or at least bound) the overall degree of the determinant as a polynomial in x.

The determinant is also the volume of the n-dimensional parallelopiped formed by the column vectors of the matrix. This is easy to check in two dimensions, for the matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}$:

To calculate the area of the central parallelogram, we have to subtract the area of two small rectangles and four small triangles from the outer rectangle, obtaining

$(a+b)(c+d)-2bc - \frac12(ac+ac+bd+bd)=ad-bc,$

as we expect. This calculation is harder to execute in higher dimensions, and certainly harder to visualise.

Maybe, though, we don’t have to, so long as we can reassure ourselves that this volume satisfies the implicit definition of the determinant map at the start. Multilinearity in the columns is not that hard to see. If we multiply the jth column by some constant, we are stretching the parallelopiped by the same constant factor in one direction, and so the volume grows appropriately. The additivity property can similarly be thought of as joining together two parallelopipeds at their common face (which is common since the other column vectors have to be constant in this construction). If two column vectors are equal, then clearly this volume actually has dimension at most n-1, and thus volume zero, so the final two conditions are genuinely easy to check.

The challenge here is that there is a direction involved. Determinants can be negative, but in our classical viewpoint, areas generally are not. In 2D, we can think of this as saying that the area is positive if the vector (b,d) lies anti-clockwise from (a,c) in the parallelogram, while is it negative otherwise. Again, this is harder to visualise in higher dimensions, but it is at least plausible that one could develop a similar decomposition. Ultimately, we are happy with the notion of directed lengths (ie vectors on the real line), and these are easy to add up without having to separate into cases, and the case holds for areas and higher-dimensional volumes.

Evaluating determinants

If we actually want to compute the determinant of a given matrix, the sum over permutations is intractable since it doesn’t have any natural splits into stages. The implicit definitions and this area consideration are clearly useless for all but the most special of examples.

The Laplace expansion is the usual algorithm to calculate the determinant of an n x n matrix. You pick a row (or a column), and evaluate the determinants of the (n-1) x (n-1) minor matrices given by deleting this row (or column), and each column (or row) in turn. This leaves us with n determinants of smaller matrices, which we pre-multiply by the entries in the original deleted row (or column), and add up in an alternating way (*). This is highly computationally intensive for large matrices, but for 3×3 and 4×4 can be done by hand with probability of an error bounded away from 1.

There is the flexibility to choose the reference row or column. Since the entries of these affect the sum through small products, it is highly convenient to choose a row or column with a lot of zeros. In particular, if there’s a row or column with exactly one non-zero entry, this is an ideal candidate.

The sum over permutations also works well when a lot of the entries are zero, because then a lot of the permutations give a summand which is zero. Upper-triangular matrices are a good example: only one permutation (the identity permutation) avoids all the zero elements underneath the diagonal.

One can also observe from the multilinearity property of the determinant map that there are lots of operations we can apply to the matrix which leave the determinant fixed. These are often called elementary row operations, though obviously we can apply them to the columns as well. To summarise, if we interchange two rows, the sign of the determinant is reversed. And if we add some multiple of one row to any other row, the determinant stays the same.

When matrices are not square, it’s quite important to be specific about exactly what form you can reduce a general matrix to via such row operations, but in this context, it’s not hugely important. Reduced echelon form (without the condition that leading coefficients will be one) is achievable, but this is a special case of an upper triangular matrix, for which the determinant is given by the product of the diagonal entries, ie is easy.

Whether this is substantially easier than Laplace expansion depends on the matrix itself and taste, both to do manually, and to code.

(*) I’m not a fan personally of this alternating definition. It seems to me much more natural to define the minor as

$M^{i,j}=(a_{i+k,j_\ell})_{1\le k,\ell\le n-1},$

with indices taken modulo n. Then you don’t have any $\pm 1$s in the Laplace expansion.

Using determinants in abstract problems

So the determinant gives directly the area of the image (under A) of the unit hypercube. By linearity (of A), it is easy to see that it also gives the scale factor of the area change (under A) of any hyper-cuboid, parallel to the conventional axes, anywhere in the space. Then, eg by approximating any sensible n-dimensional shape (*) as a union of such hyper-cuboids, we can show that in fact the area of any sensible shape increases by a factor (det A) under application of A.

This is a good thing to remember, because it is an excellent heuristic for seeing why the determinant of a linear map is basis-independent. It also gives a much easier proof of the key result

$\mathrm{det}(AB)=\mathrm{det}(A) \mathrm{det}(B),$

than that given by fixing B and viewing det(AB) as a map from matrices to the field, just like the original definition of determinant.

Some of the theory in the course is proved using elementary row operations. But these invite complicated notation, so are best used only in simple arguments, or when things are fairly explicit to begin with. Given an abstract problem about determinants of matrices, it is often tempting to induct on the size of the matrix in some way. I think it’s worth saying that even though the Laplace expansion is explicitly set up in this way, the notation involved is also likely to be annoying here, while permutations are easy to describe inductively: eg let $\sigma(1)=k$, then view the remainder of the permutation as a bijection $\{2,3,\ldots,n\}\rightarrow [n]\backslash \{k\}$.

Shortly, we’ll have a second post answering the final question: what’s the point of working with determinants? We’ve already seen half an answer, in that they describe the change-of-volume factor of a matrix (or linear map), but this can be substantially developed.

# Perturbation of Eigenvectors

In the previous post, I talked about eigenvalues, and some alternative characterisations which could be useful in some circumstances. Recently, I’ve been interested in controlling how eigenvalues and eigenvectors change as the matrix is varied. My particular example concerns positive matrices, which have a well-defined largest eigenvalue (or Perron root), and a unique (up to normalising in some way) principal eigenvector.

We might expect that perturbing a matrix slightly does not change the eigenvectors very much, since any original eigenvector is still almost an eigenvector, in the sense that its image under the action of the perturbed matrix is almost equal to a multiple of itself. But how to make this precise? And when does it go wrong?

Eigenvalues – The non-multiple case

Throughout, we assume we have a k x k matrix. We might want to allow the entries to be complex, but for now, real entries are perfectly interesting enough.

It makes sense to start with eigenvalues, since it’s easy to define these through the characteristic equation of the matrix. The coefficients of this polynomial are well-behaved (indeed polynomial) functions of the entries of the matrix. So we are really asking how the set of roots of a finite polynomial evolves as the (k+1) coefficients of the polynomial evolve. It is fairly clear that, under any sensible choice of topology on the space of k-(multi)-subsets of $\mathbb{C}$, the multiset of roots is continuous in the coefficients of the polynomial.

To say anything more precise, we have to introduce some notation.

Let $\chi_{A}(z)=z^k+\gamma_{k-1}(A)z^{k-1}+\ldots+\gamma_1(A)z+\gamma_0(A)$ be the characteristic polynomial of A. Each $\gamma_i$ is a polynomial of degree $k-i$ in the entries of A. Let’s consider now a matrix-valued function A(t), and we assume that the entries of A(t) are all differentiable with respect to t. So each $\gamma_i(A(t))$ is also differentiable with respect to t.

At this point, let’s make the assumption that t lies in some interval [r,s] for which the eigenvalues of A(t) are distinct. Let $\lambda(t)$ be some eigenvalue of A(t), chosen such that $\lambda$ is a continuous function of t. For example, we might take $\lambda(t)=\Lambda_1(t)$, the eigenvalue with largest absolute value (with some canonical tie-breaking mechanism). Then $\chi_{A(t)}(\lambda(t))=0$, and so differentiating with respect to $\gamma_i$:

$0=\chi'_{A(t)}(\lambda(t)) \frac{\partial \lambda}{\partial \gamma_i} \Big|_{A(t)} + \lambda(t)^i.$

Because we deliberately demanded that the eigenvalues were disjoint, we have $\chi'_{A(t)}(\lambda(t))\ne 0$, and so $\frac{\partial \lambda(t)}{\partial \gamma_i}=-\lambda(t)^i / \chi'_{A(t)}(\lambda(t))$. In particular, $\lambda(t)$ is differentiable with respect to the coefficients of the characteristic polynomial, and thus with respect to t also.

Multiple Eigenvalues

It gets more complicated when the characteristic equation has multiple roots. Typically we will be interested in the evolution of the eigenvalue with some extremal property, probably the largest one. Let’s restrict to the real, symmetric case, where the set of eigenvalues is complete and real. Suppose we have $t_0$ such that $A(t_0)$ has a repeated eigenvalue. Then, in a small enough region of $t_0$, we can define eigenvalues $\lambda(t),\mu(t)$ continuously such that $\lambda(t_0)=\mu(t_0)$ while $\lambda(t)\ne \mu(t)$ for $t=\ne t_0$. Then, if the entries of A(t) are analytic functions of t, then so are $\lambda(t),\mu(t)$.

But then $\max(\lambda(t),\mu(t))$ will in general not be analytic, as the maximum of two smooth functions is in general Lipschitz.

This effect is most obvious in the case of a diagonal matrix $A(t)=\begin{pmatrix}t&0\\0&-t\end{pmatrix}$, for which the largest eigenvalue is $|t|$.

Eigenvectors

When the matrix A is real and symmetric, we know it has real eigenvalues, and an orthogonal basis of eigenvectors. Then the Rayleigh quotient characterises the eigenvector as well as the eigenvalue. Recall that for any $x\in\mathbb{R}^k$ with $||x||_2=1$, we have

$\lambda_1\ge x^T A x \ge \lambda_k,$

with equality precisely at the respective eigenvectors. So if we perturb A slightly, keeping it real and symmetric, we can control the principal eigenvector quite well by this method.

If A is not diagonalisable, we can still say something about this principal eigenvector, via large powers of A, sometimes called the Van Mises iteration. This says that for large N, $A^N v$ should have direction close to that of the eigenvector, for any test vector v. The rate of convergence depends on the ratio of the largest eigenvalue to the second largest eigenvalue, though if the matrix is not diagonalisable, it is not completely trivial to quantify this convergence. We have to be careful though, since A maps the subspace orthogonal to the eigenvector to itself, so the magnitude of the projection of v onto the eigenvector determines the speed of convergence. Indeed, if v is orthogonal to the eigenvector, it won’t converge towards the principal eigenvector at all. (But if there is a well-defined ‘second eigenvector’ then it will converge towards that.)

Continuity of Eigenvectors

The reason why I ended up reading about some of these topics was that I wanted to show that the Perron eigenvector of a positive matrix (that is, the unique eigenvector corresponding to the Perron root) was Lipschitz continuous as a function of the entries of the positive matrix. Since for such a matrix, the largest eigenvalue is simple, we are able to make some progress.

In general, the condition that $v$ is an eigenvector of matrix A with eigenvalue $\lambda$ is described by the relation:

$(A-\lambda I)v=0,\quad ||u||_1=1,$ (*)

or whatever the most appropriate normalising condition appears. This describes an implicit relation between A and the eigenvalue-eigenvector pair $(\lambda,v)$. So given a matrix $A_0$ with eigenvalue $\lambda_0$ corresponding to eigenvector $v_0$, in a neighbourhood of $A_0$ in $\mathbb{R}^{k\times k}$ we can use the implicit function theorem to comment on the differentiability of $(\lambda,v)$ with respect to A in this neighbourhood.

Precisely, we require the matrix of partial derivatives from (*)

$\begin{pmatrix} A_0-\lambda_0 I&v_0 \\ \mathbf{1}&0\end{pmatrix}$

to have non-zero determinant. But if $\lambda_0$ is not simple, then if we apply this matrix from the left to one of the other eigenvectors (with a zero appended) we can see that it has non-trivial kernel. With a bit more work, we can show the converse too, and conclude that $(\lambda,v)$ are smooth with respect to A in some neighbourhood of $A_0$.

Finally, we observe that when the eigenvalues are not simple, we can’t even guarantee continuity of the eigenvectors. This is unsurprising really, since for a multiple eigenvalue, a) we might not know how many LI eigenvectors exists; and b) we might have complete freedom over the choice of eigenvectors. Think about the identity matrix! Indeed the eigenvectors of $\begin{pmatrix}1+\epsilon &0\\ 0&1+\epsilon\end{pmatrix}$ are (1,0) and (0,1), while the eigenvectors of $\begin{pmatrix}1&\epsilon\\ \epsilon&1\end{pmatrix}$ are (1,1), (1,-1). So no continuous choice of eigenvectors is possible here.

# Characterisations of Eigenvalues

I’ve been working for much of the past few months on a version of the frozen percolation random graph process with types. The connectivity between types is controlled by a (finite) non-negative square matrix, and so I’ve been engaging with linear algebra theory to an extent I haven’t really experienced since the second or third year of undergraduate maths.

We are interested in whether the graphs in question are subcritical, critical or supercritical. As in the case of multitype branching processes, this is controlled by the principal eigenvalue of a related non-negative matrix. So I’ve been looking up lots of methods for controlling eigenvalues, and some have proved useful, and some have not, but I thought it would be worthwhile to present some of them here.

Bounds and characterisations of spectral radius

Throughout, I will be talking about finite, square matricies. Eigenvalues may be defined as roots of the characteristic polynomial, and so by the fundamental theorem of algebra, there is always at least one complex eigenvalue. There is always at least one eigenvector associated to any eigenvalue. However, the dimension of the eigenspace is not always the same as the multiplicity of the eigenvalue as a root of the characteristic polynomial. The latter is called algebraic multiplicity, while the former is geometric multiplicity.

For now though, this distinction will be unimportant. The spectral radius of a matrix A is defined as

$\rho(A)=\max \{|\lambda|\, : \, \lambda \text{ and eigenvalue of }A\}.$

We can bound the spectral radius in terms of the norm of the matrix. Remember that a matrix norm has to satisfy all the usual properties of a norm, as well as a submultiplicative property $|||AB|||\le |||A|||\cdot |||B|||$. This is good, as otherwise we would be free to replace any norm by an arbitrary multiple of itself, and so no useful bounds could ever emerge. Note that the submultiplicativity implies that $|||I_n||\ge 1$.

Now, let $\lambda,x$ be some eigenvalue and associated (right-)eigenvector respectively of matrix A. Let X be the square matrix given by taking all the columns to be x. Now $Ax=\lambda x$ implies $AX=\lambda X$, and so

$|\lambda| \cdot|||X||| = |||\lambda X||| = |||AX||| \le |||A|||\cdot |||X|||,$

and thus we conclude our most basic bound $\lambda \le |||A|||$.

When A is diagonalisable, life is particularly easy, but in general we can write A as a conjugate of its Jordan normal form. Then, by looking at each diagonal block of the Jordan normal form separately, we can show that

$\lim_{k\rightarrow 0}A^k = 0\quad \iff \quad \rho(A)<1.$

Then, applying this, with additional care, to the matrices $A / (\rho(A)\pm \epsilon)$, we derive Gelfand’s Formula, that $\rho(A) = \lim_{k\rightarrow \infty} ||A^k||^{1/k}$. Again, this applies for any matrix norm.

Real symmetric matrices

When the matrix is real and symmetric, it is not too hard to show that all the eigenvalues are real, and furthermore that all the geometric multiplicities are equal to the algebraic multiplicities. That is, the matrix is diagonalisable, and there is an (orthogonal) basis of eigenvectors. Once we assume we are working with respect to this eigenbasis, it is easy to see how the Rayleigh quotient characterisation of the largest (and smallest) eigenvalue works. Let’s say the eigenvalues are $\lambda_1\ge \lambda_2\ge\ldots \ge \lambda_n$, then for any $||x||_2=1$, we have $\lambda_1\ge x^T A x\ge \lambda_n$, and equality is attained when x is the respective eigenvector, normalised appropriately.

This is an especially useful characterisation of the largest eigenvalue, as for example we can see fairly easily that this means $\lambda_1$ is a convex function of the (real, symmetric) matrix.

We can generalise this Rayleigh quotient idea if we take k orthonormal vectors in $R^k$, arrange them in an nxk matrix P, so that $P^T P = I_k$. Now we consider the matrix $P^TAP$. [Note that if k=1, we are exactly considering $x^TA x$ as before.] Then Poincare’s Separation Theorem say that the eigenvalues $\mu_1\ge \mu_2\ge\ldots \mu \mu_k$ of $P^TAP$ (which is also real, symmetric) are bounded by the original eigenvalues:

$\lambda_{n-k+i} \ge \mu_i\ge \lambda_i.$

Since the trace is preserved under conjugation, and the trace is the sum of eigenvalues, we can apply this result with P’s columns taken to be the any k canonical basis vectors of $\mathbb{R}^k$. Without loss of generality, we may assume the basis has been chosen so that the diagonal elements of A satisfy $a_{11}\ge a_{22}\ge\ldots\ge a_{nn}$, and so now we have that the sequence $(a_{11},a_{22},\ldots,a_{nn})$ is majorised by $(\lambda_1,\lambda_2,\ldots,\lambda_n)$ and majorises $(\lambda_n,\lambda_{n-1},\ldots,\lambda_1)$. The first of these relations can be used via the setup of Karamata’s inequality to conclude that for any convex function f, we have

$\sum_{i=1}^n f(\lambda_i)\ge \sum_{i=1}^n f(a_{ii}).$

Gershgorin Circles

In fact, we can relate the eigenvalues to the diagonal entries of the matrix in a more general setting. We are motivated by the thought that if the off-diagonal entries are all very small, then the set of eigenvalues should be approximately given by the set of diagonal entries.

For a square complex matrix, let $\lambda,x$ be an eigenvalue, eigenvector pair. For any index i, we have

$\lambda - a_{ii}= \frac{\sum_j a_{ij}x_j}{x_i} - a_{ii} = \frac{\sum_{j\ne i}a_{ij}x_j}{x_i}.$

Now consider the i such that $x_i=\max |x_j|$, and take absolute values and apply the triangle inequality,

$|\lambda - a_{ii}| \le \sum_{j\ne i} \left| \frac{a_{ij}x_j}{x_i} \right| \le \sum_{j\ne i}|a_{ij}|.$

Let’s define $R_i=\sum_{j\ne i}|a_{ij}|$ to be the sum of the non-diagonal entries of the ith row. Then the Gershgorin circle theorem says that every eigenvalue lies within at least one of the discs $B(a_{ii},R_i)$, in the complex plane. So our motivation still makes sense. If the off-diagonal entries are small, this is a strong restriction, and if they are not typically smaller than the diagonal entries, then we perhaps do not learn very much. Obviously, we could apply the same argument to the columns too.

When the diagonal entries are distinct, and the off-diagonal entries are small, the Gershgorin discs are distinct, and we would expect each to contain exactly one eigenvalue, corresponding to the appropriate diagonal entry. In fact, we can say something stronger. In general, the union of the discs is a subset of the complex plane with some connected components. Then, if a component is the union of exactly r discs, then it contains exactly r of the eigenvalues.

To see this, consider multiplying all the off-diagonal entries by $z\in[0,1]$ and observe what happens as z varies from 0 to 1. When z=0, the matrix is diagonal, and each eigenvalue is in the Gershgorin disc (which is a single complex number). As z varies continuously, the characteristic polynomial varies continuously, and also its roots, that is the set of eigenvalues. So since each of the r eigenvalues are initially within the union of the r original, large Gershgorin discs, they must remain within this union as z varies, since they cannot ‘jump’ to another component.

It’s hard to know how time will allow, but provisionally in the next post I will talk about how to control the evolution of eigenvectors as a function of the matrix, and in particular what can go wrong.

REFERENCES

For the middle section, I used the progression from Chapter 4 of Matrix Differential Calculus with Applications in Statistics and Econometrics (Magnus and Neudecker).

# Duality for Interacting Particle Systems

Yesterday I introduced the notion of duality for two stochastic processes. My two goals for this post are to elaborate on the idea of why duality is useful, which we touched on in passing in the previous part, and to discuss duality of interacting particle systems. In the latter case, there are often nice ways to consider the forward and backward processes together that make the relation somewhat more natural.

The starting point is to assume a finite state space. This will be reasonable when we start to consider interacting particle systems, eg on $\{0,1\}^{[n]}$. As before, call the spaces R and S, and a duality function H(x,y). Since the state-spaces are finite, it is entirely natural to think of this as a matrix, and hence as an operator. Of course, a function defined on a finite state-space can be thought of as a vector, so it is clear what this operator will actually operate on. (I’ve chosen H rather than h for the duality function so it is more clear that it is acting as an operator here.)

We have some choice about which way round to define it, but for now let’s say that given some function f(.) on S

$Hf(x):=\sum_{y\in S} H(x,y)f(y).$

Note that this is a) exactly the definition of matrix (left-)multiplication; b) We should think of Hf as a function on R – perhaps (Hf)(x) might be more clear? and c) the operator H acts $\mathbb{R}^S\rightarrow \mathbb{R}^R$. If we want the corresponding operator $\mathbb{R}^R\rightarrow\mathbb{R}^S$, we simply multiply by H on the right instead.

But note also that the generator of a finite state-space Markov process is also a matrix, indeed a Q-matrix. So if we take our definition of the duality function as

$\mathcal{G}_X h(x,y)=\mathcal{G}_Y h(x,y),$

which, importantly, holds for all x,y, we can convert this into an algebraic form as

$\mathcal{G}_X H = H \mathcal{G}_Y^\dagger.$

In the same way that n-step transition probabilities for a discrete-time Markov chain are given by the product of the one-step transition matrix, general time transition probabilities for a continuous-time Markov chain are given by exponents of the Q-matrix. In particular, if X and Y have transition kernels P and Q respectively, then $P_t=e^{tG_X}$, and after doing some manipulation, we can show that

$P_t H=H Q_t^\dagger,$

also. This is really useful as in general we would hope that H might be invertible, from which we derive

$P_t=HQ_t^\dagger H^{-1}.$

So this is a powerful statement about the relationship between the evolutions of the two processes. In particular, it shows a correspondence (given by H) between left eigenvectors of P, and right eigenvectors of Q, and vice versa naturally.

The reason why this is useful rather than merely cute, is that when we re-interpret everything in terms of the original stochastic processes, we get a map between stationary distributions of X, and harmonic functions of Y. Stationary distributions are often hard to describe in any terms other than the left-1-eigenvector, or through some convergence property that is typically hard to work with. Harmonic functions, on the other hard, can be much more tractable. An example of a harmonic function is the survival probability started from a given state. This is useful for specifying the stationary distribution, but perhaps even more so for describing properties of the set of stationary distributions. In particular, uniqueness and existence are carried across this equivalence. So, for example, if the dual does not survive almost surely, then this says the only stationary measure is zero, and so the process is transient or similar.

Jan Swart’s course in Luminy last October dealt with duality, with a focus mainly on interacting particle systems. There are a couple of themes I want to talk about, without going into too much detail.

A typical interacting particle system will take place on a locally finite graph. At each vertex, there is either a particle, or there isn’t. Particles move between adjacent vertices, and sometimes interact with particles at adjacent vertices. These interactions might involve branching or coalescence. We will discuss shortly the set of possible forms such interaction might take. The state space is $\{0,1\}^{V(G)}$, with G the underlying graph. Then given a state, there is some set of actions which might happen next, and we consider the possibility that they happen with exponential rates.

At this stage, it seems like the initial configuration is important, as this affects what set of moves can happen immediately, and also thereafter. It is not clear how quickly this dependence fades. One useful idea is not to restrict ourselves to interactions involving the particles currently present in the system, but instead to consider a Poisson process of all possible interactions. Only the moves actually permitted by the current state will happen, but having this extra information allows for coupling between initial configurations.

It’s probably easier to consider a concrete example. The picture below shows the set-up for a branching random walk up an integer lattice. Each particle moves to one of the two state directly above its current state, or it branches and sends particles to both of them.In the diagram, we have glued arrows onto every state at every time, which tells us what to do if there is a particle there at each time. As a coupling, we can now think of the process as a deterministic walk through a random environment. The environment is given by some probability space, which in continuous time might have the appearance of a Poisson process on the set of ‘moves’, and the initial condition of the walk is up to us.

We can generalise this to a broader class of interacting particle systems. If we want all interactions to be between pairs of adjacent states, there are six possible things which could happen:

• Annihilation: two adjacent particles destroy each other. ( 11 -> 00 )
• Branching: one particle becomes two particles. ( 01 or 10 -> 11 )
• Coalescence: two particles merge. ( 11 -> 01 or 10 )
• Death: A particle is removed. ( 01 or 10 -> 00 )
• Exclusion: a particle moves. ( 01 -> 10 )
• Birth: a particle is created. ( 00 -> 01 or 10 )

For now we exclude the possibility of birth. Note that the way we have set this up involving two-site interactions excludes the possibility of a particle trying to move to an already-occupied site.

Let us say that in process X the rates at which each of these events happen are a, b, c, d and e, taking advantage of the helpful choice of naming. There is some flexibility about whether the rates are the same between every pair of vertices of note. For this post we assume that they are. Then it is a result of Lloyd and Sudbury that given some real $q\neq 1$, the process X’ with corresponding rates given by:

$a'=a+2q\gamma, b'=b+\gamma, c'=c-(1+q)\gamma, d'=d+\gamma, e'=e-\gamma,$

for $\gamma:= \frac{a+c-d+qb}{1-q},$

is dual to X, with duality function given by $h(Y,Z)=q^{|Y\cap Z|}$, for Y and Z possible states.

I want to make two comments:

1) This illustrates one of the differences between the dual and the time-reversal. It is clear that the time-reversal of branching is coalescence and vice versa, and exclusion is invariant under time-reversal. But the time-reversal of death is definitely birth, but there is no birth component in the dual of a process which features death. I don’t have a strong intuition for why this is the case, but see the final paragraph of this post. However, at least it seems plausible that both processes might simultaneously be recurrent, since in the dual, both the branching rate and the death rate have increased by the same amount.

2) This settles one problem of uniqueness of the dual that I mentioned last time, since we can vary q and get a different dual to the same original process. For example, in the voter model, we have b=d=1, and a=c=e=0, as in any update, the opinions of neighbours which were previously different become the same. Anyway, for any $q\in[-1,0]$ there is a choice of dual, where at the extremes q=0 corresponds to coalescing random walk, and q=-1 to annihilating random walk. (Note that the duality function for q=0 is the indicator function that the systems are different.)

As a final observation without much justification, suppose we add in arrows in the gaps of the branching random walk picture we had earlier, and direct them in the opposite direction. It turns out that this corresponds precisely to the dual of the process. This provides an appealing visual idea of why the dual of branching might be death. It also supports the general idea based on the coupling described earlier that the dual process is in some sense a deterministic walk in the opposite direction through the random environment specified by the original process.

REFERENCES

J.M. Swart – Duality and Intertwining of Markov Chains (mainly using chapters 2.1 and 2.7)

Thanks for Daniel Straulino for direction towards the branching random walk duality example.

# Convergence of Transition Probabilities

As you can see, I haven’t got round to writing a post for a while. Some of my reasons for this have been good, and some have not. One reason has been that I’ve had to give a large number of tutorials for the fourth quarter of the second year probability course here in Oxford. The second half of this course concerns discrete-time Markov chains, and the fourth problem sheet discusses various modes of convergence for such models, as well as a brief tangent onto Poisson Processes. I’ve written more about Poisson Processes than perhaps was justifiable in the past, so I thought I’d say some words about convergence of transition probabilities in discrete-time Markov chains.

Just to be concrete, let’s assume the state space K is finite, and labelled {1,2,…,k}, so that it becomes meaningful to discuss

$p_{12}^{(n)}:=\mathbb{P}(X_n=2|X_0=1).$

That is, the probability that if we start at state 1, then after n ‘moves’ we are at state 2. We are interested in the circumstances under which this converges to the stationary distribution. The heuristic is that we can view a time-step of a Markov chain as an operation on the space of distributions on K. Note that this operation is deterministic. If this sounds complicated, what we mean is that we specify an initial distribution, that is the distribution of $X_0$. If we consider the distribution of $X_1$, this is given by $\lambda P$, where $\lambda$ is the initial distribution, and P the transition matrix.

Anyway, the heuristic is that the stationary distribution is the unique fixed point of this operation on the space of distributions. It is therefore not unreasonable to assume that unless there are some periodic effects, we expect repeated use of this operation to move us closer to this fixed point.

We can further clarify this by considering the matrix form. Note that a transition P always has an eigenvalue equal to 1. This is equivalent to say that there is a solution to $\pi P=\pi$. Note it is not immediately equivalent to saying that P has a stationary distribution, as the latter must be non-negative and have elements summing to one. Only the first property is difficult, and relies on some theory or cleverness to prove. It can also be shown that all eigenvalues satisfy $|\lambda|\le 1$, and in general, there will be a single eigenvalue (ie dimension 1 eigenspace) with $|\lambda|=1$, and the rest satisfies $|\lambda|<1$. Then, if we diagonalise P, it is clear why $\pi P^n$ converges entry-wise, as $\pi UP^n U^{-1}$ converges. In the latter, only the entries in the row corresponding to $\lambda=1$ converge to something non-zero.

In summary, there is a strong heuristic for why in general, the transition probabilities should converge, and if they converge, that they should converge to the stationary distribution. In fact, we can prove that for any finite Markov chain, $p_{ij}^{(n)}\rightarrow \pi_j$, provided we two conditions hold. The conditions are that the chain is irreducible and aperiodic.

In the rest of this post, I want to discuss what might go wrong when these conditions are not satisfied. We begin with irreducibility. A chain is irreducible if it has precisely one communicating class. That means that we can get from any state to any other state, not necessarily in one step, with positive probability. One obvious reason why the statement of the theorem cannot hold in this setting is that $\pi$ is not uniquely defined when the chain is not irreducible. Suppose, for example, that we have two closed communicating classes A and B. Then, supported on each of them is an invariant distribution $\pi^A$ and $\pi^B$, so any affine combination of the two $\lambda \pi^A+(1-\lambda) \pi^B$ will give a stationary distribution for the whole chain.

In fact, the solution to this problem is not too demanding. If we are considering $p_{ij}^{(n)}$ for $i\in A$ a closed communicating class, then we know that $p_{ij}^{(n)}=0$ whenever $j\not\in A$. For the remaining j, we can use the theorem in its original form on the Markov chain, with state space reduced to A. Here, it is now irreducible.

The only case left to address is if i is in an open communicating class. In that case, it suffices to work out the hitting probabilities starting from i of each of the closed communicating classes. Provided these classes themselves satisfy the requirements of the theorem, we can write

$p_{ij}^{(n)}\rightarrow h_i^A \pi^A_j,\quad i\not\in A, j\in A.$

To prove this, we need to show that as the number of steps grows to infinity, the probability that we are in closed class A converges to $h_i^A$. Then, we decompose this large number of steps so to say that not only have we entered A with roughly the given probability, but in fact with roughly the given probability we entered A a long time in the past, and so there has been enough time for the original convergence result to hold in A.

Now we turn to periodicity. If a chain has period k, this says that we can split the state space into k classes $A_1,\ldots,A_k$, such that $p_{ij}^{(n)}=0$ whenever $n\not\equiv j-i \mod k$. Equivalently, the directed graph describing the possible transitions of the chain is k-partite. This definition makes it immediately clear that $p_{ij}^{(n)}$ cannot converge in this case. However, it is possible that $p_{ij}^{(kn)}$ will converge. Indeed, to verify this, we would need to consider the Markov chain with transition matrix $P^k$. Note that this is no longer irreducible, as it there are no transitions allowed between classes $A_1,\ldots,A_k$. Indeed, a more formal definition of the period, in terms of the lcd of possible return times allows us to conclude that there is no finer reducibility structure. That is, $A_1,\ldots,A_k$ genuinely are the closed classes when we consider the chain with matrix $P^k$. And so the Markov chain with transition matrix $P^k$ restricted to any of the $A_i$s satisfies the conditions of the theorem.

There remains one case which I’ve casually brushed over. When we were discussing the irreducible case, I said that if we had at least one communicating classes, then we could work out the limiting transition probabilities from a state in an open class to a state in a closed class by calculating the hitting probability of that closed class, then applying the standard version of the theorem to that closed class. This relies on the closed class being aperiodic.

Suppose otherwise that the destination closed class A has period k as before. If it were to be the case that the number of steps required to arrive at A had some fixed value mod k, or modulo a non-trivial divisor of k, then we certainly wouldn’t have convergence, for the same reasons as in the globally periodic case. However, we should ask whether we can ever have convergence?

In fact, the answer is yes. For concreteness, and because it’s easier to write ‘odd’ and ‘even’ than $m \mod k$, let’s assume A has size 2 and period 2. That is, once we arrive in A, thereafter we alternate deterministically between the two states. Anyway, for some large time n, we can write $p_{ca}^{(n)}$ for $a\in A, c\not\in A$ as:

$p_{ca}^{(n)}=h_i^A(n),$

where the latter term is the probability that we arrive in A at a time-step which has the same parity as n. It’s not terribly hard to come up with an example where this holds, and this idea holds in greater generality, where A has period k (and not necessarily just k states), we have to demand that the probability of arriving at a time which is a mod k is equal for all a in [0,k-1].

Of course, for applications, we don’t normally care much about irreducible chains, and we can easily remove periodicity by introducing so-called laziness, whereby on each time-step we flip a coin (biased if necessary) and stay put if it comes up heads, and apply the transition matrix if it comes up tails. Then it’s possible to get from any state to itself in one step, and so we are by construction aperiodic.

# The Perron-Frobenius Theorem for Stochastic Matrices

This article was prompted by a question asked by one of my students about 5 minutes before the end of the final lecture of my Markov Chains course in Linyi. Although I didn’t have time to think about it right then, the flight home offered plenty of opportunity for having a play around. I am well aware that the proof given below is not the best one, but the ideas of minimising quadrants of a matrix seemed fairly natural. Anyway, it’s been sitting on my desktop for over two months now, so I decided I should put it up.

———

Recall that the transition probabilities of a finite Markov chain are determined by a stochastic matrix P. That is, each row of P is a probability distribution. In both theory and applications, the existence and properties of an invariant distribution is of interest. This is a probability distribution $\pi$ satisfying the relation:

$\pi P=\pi.$ (1)

It is clear that $\bigg(\begin{smallmatrix}1\\ \vdots\\1\end{smallmatrix}\bigg)$ is a right- or column eigenvector of P, with eigenvalue 1. Since the spectrum of $P^T$ is the same as that of P, we conclude that 1 is a left-eigenvalue of P also. So we can be assured of the existence of a vector $\pi$ satisfying (1). What is unclear is that this eigenvector $\pi$ should be a probability distribution. Since at least one entry must be non-zero, it will suffice to show that every entry of $\pi$ is non-negative.

A necessary condition for the uniqueness of an invariant distribution is that the Markov chain be irreducible. This is best defined using the terminology of random walks: the chain is irreducible if for every pair of states $i,j\in I$, it is possible to move from i to j and back again. In terms of the transition matrix, P is irreducible if it is not block upper-triangular, up to reordering rows and columns.

We want to show that when P is irreducible, the (unique) 1-eigenvector is a probability distribution. The standard method proposed in this context is to exhibit the invariant distribution directly. For example, Norris’s Markov Chains defines

$\gamma_i^k=\text{ expected time spent in i between visits to k }=\mathbb{E}_k\sum_{n=0}^{T_k}1_{\{X_n=i\}},$

and shows that $(\gamma_i^k)_{i\in I}$ satisfies (1).

Nonetheless, the result required is clearly at least one step removed from the probabilistic interpretation, so it would be satisfying to find a direct proof of existence. Typically, one quotes the substantially more general theorem of Perron and Frobenius, the most relevant form of which is:

Theorem (Perron-Frobenius): Given A a non-negative and irreducible square matrix. Then there is a positive real eigenvalue $\lambda$ with multiplicity 1 and such that all other eigenvalues have absolute value less than or equal to $\lambda$. Then the (unique up to scaling) left- and right-eigenvectors corresponding to $\lambda$ are positive.

Here we present a short proof of this result in the case where A is the (stochastic) transition matrix of a Markov chain.

Proposition: An irreducible stochastic matrix has a 1-eigenvector with all entries non-negative.

Proof: We show instead the contrapositive: that if a stochastic matrix has a 1-eigenvector with both negative and non-negative components, then it is reducible. The motivation is this third eigenvector given in example (2). Observe that the communicating classes are $\{1,2\}$ and $\{3\}$, and it is not hard to see that for any eigenvector with negative and non-negative components, the sign of a component is a class property.

Informally, given an $n\times n$ stochastic matrix P, and a 1-eigenvector $\pi$ with this property, we relabel the states so that the non-negative components, which we call $A\subset I$ are first. That is, in a moderate abuse of notation:

$\pi=(\underbrace{\pi_A}_{\geq 0}\quad\underbrace{\pi_B}_{<0}).\quad\text{ If we write P as }\begin{pmatrix}P_{AA}&P_{AB}\\P_{BA}&P_{BB}\end{pmatrix},$

our aim is to show that the sub-matrices $P_{AB}$ and $P_{BA}$ are both zero. This implies that states in A and states in B do not communicate, showing that P is reducible. We can formulate this as a linear programming problem:

$\text{Maximise }\sum_{\substack{x\in A,y\in B\\x\in B, y\in A}}p_{xy}\text{ s.t. }\begin{cases}p_{xy}\geq 0&\forall x,y\in I\\p_{x1}+\ldots+p_{xn}=1&\forall x\in I\\\pi_1p_{1y}+\ldots+\pi_np_{ny}=\pi_y&\forall y\in I\end{cases}$

It suffices to show that this maximum is 0. Now, we take $|A|=i$, and assume that $1\leq i\leq n-1$, that is, there are a positive number of negative and non-negative components. Noting that the sum of the rows in a stochastic matrix is 1, we may consider instead:

$\text{Minimise }\sum_{\substack{x,y\in A\\x,y\in B}}p_{xy}\text{ s.t. }\begin{cases}p_{xy}\geq 0&\forall x,y\in I\\p_{x1}+\ldots+p_{xn}=1&\forall x\in I\\\pi_1p_{1y}+\ldots+\pi_np_{ny}=\pi_y&\forall y\in I\end{cases}$

and it will suffice to show that this minimum is n. To do this, we consider instead the dual:

$\text{Maximise }\lambda_1+\ldots+\lambda_n+\pi_1\mu_1+\ldots+\pi_n\mu_n,$

$\text{ s.t. }\lambda_x+\pi_y\mu_x\leq\begin{cases}1&\text{if }x,y\leq i\text{ or }x,y\geq i+1\& \text{otherwise}\end{cases}$

The objective is certainly bounded by n. And in fact this is attainable, for example by taking:

$\lambda_1=\ldots=\lambda_i=1,\quad \lambda_{i+1}=\ldots=\lambda_n=0$

$\mu_1=\ldots=\mu_i=0,\quad \mu_{i+1}=-\frac{1}{\pi_{i+1}}, \ldots,\mu_n=-\frac{1}{\pi_n}.$

Applying strong duality for linear programming problems completes the proof.

# Invariant Distributions of Markov Chains

My lecture course in Linyi was all about Markov chains, and we spent much of the final two sessions discussing the properties of invariant distributions. I was not surprised, however, that none of the class chose this topic as the subject for a presentation to give after the end of the teaching week. One of the main problems is that so many rather similar properties are introduced roughly simultaneously. As we did in the class, I thought it was worth making some sort of executive summary, as a mixture of revision and interest.

Definition: $\pi$ is an invariant measure if $\pi P=\pi$. If in addition $\sum_{i\in I}\pi_i=1$, then we say it is an invariant distribution. Of course, if I is finite, then any invariant measure can be normalised to give an invariant distribution.

The key initial questions are about existence and uniqueness. First, if there are multiple communicating classes, then an invariant measure (resp. distribution) is a linear (resp. affine) combination of the invariant measures / distributions on each (closed) class. So we restrict attention to irreducible Markov chains.

In the finite case, P is a stochastic matrix so has a column eigenvector with eigenvalue 1, namely the vector with all entries equal to 1. Thus, by reference to general theory in linear algebra, P has a row eigenvector $\pi$ with eigenvalue 1. To paraphrase a remark made by one of my students, what is not clear is that this should be a measure. Demonstrating that this is true is rather non-trivial I think, normally done by reference to the rather more general Perron-Frobenius theorem, though on the flight home I came up with a short argument using Lagrangian duality. For now, we accept existence in the finite case, and note that we typically show existence by showing that the vector of expected time spent in each state between successive visits to a fixed reference state satisfies the properties of an invariant measure.

This is a good moment to note that recurrence is not a necessary condition for the existence of an invariant measure. For example, the random walk on $\mathbb{Z}^3$ is transient, but the uniform measure is invariant. However, it is not a sufficient condition for the existence of an invariant distribution either. (Of course, an irreducible finite chain is always recurrent, and always has an invariant distribution, so now we are considering only the infinite state space case.) The random walk on $\mathbb{Z}^2$ is recurrent, but the invariant measure is not normalisable.

The property we in fact need is positive recurrence. This says that the expected return time to each point is finite. Again, this is a class property. This is a common requirement in probabilistic arguments: almost surely finite is often not strong enough to show results if the expectation is infinite (see for example the various requirements for the optional stopping theorem). If this holds, then $\pi_i=\frac{1}{\mathbb{E}T_i}$, where $T_i$ is the the return time starting from some $i\in I$.

The final question is ‘Why are we interested?’ One of the best answers is to look at convergence properties. A simple suggestion is this: if we start in equilibrium, then $X_0,X_1,X_2,\ldots$ are all equal in distribution. Note that the dependence structure remains complicated, and much much more interesting than the individual distributions. Next, we observe that a calculation of n-step transition probabilities for a finite chain will typically involve a linear combination of nth powers of eigenvalues. One of the eigenvalues is 1, and the others lie strictly between -1 and 1. We observe in examples that the constant coefficient in $p_{ij}^{(n)}$ is generally a function of j alone, and so $p_{ij}^{(n)}\rightarrow\lambda_j$, some distribution on I. By considering $P^{n+1}=P\cdot P^n$, it is easy to see that if this converges, $(\lambda_j)_{j\in I}$ is an invariant distribution. The classic examples which do not work are

$P=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ and $P=\begin{pmatrix}0&1&0\\ 0&0&1\\1&0&0\end{pmatrix}$,

as then the distribution of $X_n$ is a function of the remainder of n modulo 3 alone. With a little thought, we can give a precise classification of such chains which force you to be in particular proper subsets of the state space at regular times n. Chains without this property are called aperiodic, and we can show that distributions for such chains converge to the equilibrium distribution as $n\rightarrow\infty$.