# 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.

# Discontinuous Phase Transitions

Yesterday, Demeter Kiss from Cambridge gave a seminar in Oxford about a model for self-destructive percolation on $\mathbb{Z}^2$ that had implications for the (non-)existence of an infinite-parameter forest fire model on the same lattice. I enjoyed talking about this and his recent work on the related model of frozen percolation on $\mathbb{Z}^2$. Considering these models in the lattice setting present a whole range of interesting geometric challenges that are not present in the mean-field case that has mainly occupied my research direction so far.

The afternoon’s discussion included lots of open problems about percolation. Several of these are based around continuity of the phase transition, so I thought I would write a quite post about some simple examples of this, and one example where it does not hold.

A helpful base example is bond percolation on the lattice $\mathbb{Z}^2$. Here, we specify some probability p in [0,1], and we declare edges of the lattice open with probability p, independently of each other. We then consider the graph induced by the open edges. We say that percolation occurs if the origin is contained in an infinite open component. The terminology arises from the interpretation as fluid being added at the origin and flowing down open edges. We define $\theta(p)$ to be the probability that the origin is in an infinite component when the parameter is p. By translation-invariance, we can get some sort of 0-1 law, to conclude that there is an infinite component somewhere in the system with probability either 0 or 1, depending on whether $\theta(p)$ is positive or zero. Indeed, we can further show that if it is positive, then with probability 1 there is a unique infinite component.

We define the critical probability $p_c:= \inf\{\theta(p)>0\}$. A question worth asking is then, what is $\theta(p_c)$? In some examples, we can find $p_c$, but we cannot prove that $\theta(p)$ is continuous around $p_c$. In the case of $\mathbb{Z}^2$ this is known, and it is known from work of Kesten that $p_c=1/2$. See below for a plot of $\theta(p)$ in this setting (obtained from this blog, though possibly originating elsewhere).

The aim is to find an example where we do not have such a continuous phase transition. The original work on frozen percolation took place on trees, and one of Kiss’s results is confirms that these show qualitatively different phenomena to the same process on the lattice. In some sense, trees lie halfway between a lattice and a mean-field model, since there is often some independence when we look down the tree from a given generation, if it is well-defined to use such language.

Anyway, first we consider percolation on an infinite regular rooted k-ary tree. This means we have a root, which has k children, each of which in turn has k children, and so on. As before we consider bond percolation with parameter p. In this setting, we have a language to describe the resulting open component of the root. The offspring distribution of any vertex in the open component is given by Bin(k,p) independently of everything else, so we can view this component as the realisation of a Galton-Watson tree with this offspring distribution. This distribution has finite mean kp, and so we can state explicitly when the survival probability is positive. This happens when the mean is greater than 1, ie p>1/k.

For our actual example, we will consider the survival probability, but the technicalities are easier to explain if we look at the extinction probability, now using the language of branching processes. Suppose the offspring distribution has pgf given by

$f(x)=p_0+p_1x+p_2x^2+\ldots.$

Then the extinction probability q satisfies f(q)=q. I want to pause to consider what happens if this equation has multiple solutions. Indeed, in most interesting cases it will have multiple solutions, since f(1) will always be 1 if it is a non-defective offspring distribution. It is typically cited that: the extinction probability q is the smallest solution to this equation. I want to discuss why that is the case.

To approach this, we have to consider what extinction means. It is the limit in the event sense of the events {we are extinct after n generations}. Let the probabilities of these events be $q_n$, so $q_0=0$. Then by a straightforward coupling argument, we must have

$0=q_0\le q_1\le q_2 \le\ldots\le q:= \lim q_n \le 1.$

But, by the same generating function argument as before, $q_{n+1}=f(q_n)\ge q_n$. So if we split [0,1] into regions A where $f(x)\ge x$ and B where $f(x), all the $(q_n)$s must occur in the former, and so since it is closed, their limit must be in A also. Note that if f(x) intersects x lots of times, then region A is not necessarily connected. In the diagram below, in moving from $q_n$ to $q_{n+1}$ we might jump across part of B.

This is bad, as we are trying to prove that q is the right boundary of the connected component of A containing 0. But this cannot happen, as f is monotonic. So if one of the roots of f(x)=x in between the hypothesised $q_n is called z, then $f(q_n)< f(z)=z < q_{n+1}$, a contradiction.

Ok, so now we are ready to consider our counterexample to continuity over the percolation threshold. See references for a link to the original source of this example. We have to choose a slightly more complicated event than mere survival or extinction. We consider bond percolation as before on the infinite ternary tree, where every vertex has precisely 3 offspring. Our percolation event is now that the root is the root of an infinite binary tree. That is, the root has at least two children, each of which have at least two children, each of which, and so on.

If we set this probability equal to q, and the probability of an edge being open equal to p, then we have the recurrence:

$q=3p^2(1-p)q^2+p^3[3q^2(1-q)+q^3].$

The first term corresponds to the root having two open edges to offspring, and the second to the root having all three open edges to offspring. After manipulating, we end up with

$q\left[2p^3q^2-3p^2q+1\right]=0.$

We are therefore interested in roots of the quadratic lying between 0 and 1. The discriminant can be evaluated as

$\Delta=p^3(9p-8),$

and so there are no real roots where p<8/9. But when p=8/9, we have a repeated root at q=27/32, which is obviously not zero!

This equation is qualitatively different to the previous one for the extinction probability of a Galton-Watson tree. There, we had a quadratic, with one root at 1. As we varied p, the other root moved continuously from greater than one to less than one, so it passed through 1, giving continuity at the critical probability. Here, we have a cubic, again with one root at 1. But now the other roots are complex for small p, meaning that the local minimum of the cubic lies above the x-axis. As p gets to the critical value, it the local minimum passes below the x-axis, and suddenly we have a repeated root, not at zero.

I would like to have a neat probabilistic heuristic for this result, without having to make reference to generating functions. At the moment, the best I can come up with is to say that the original problem is simple, in the sense that the critical probability is as small as it could be while still making sense in expectation. To be concrete, when the mean of the offspring generation is less than 1, the expected size of the nth generation tends to zero, so there certainly could not be positive probability of having an infinite component.

Whereas in the binary tree example, we only require p=2/3 to have, in expectation, the right number of open edges to theoretically allow an infinite binary tree. If we think of percolation as a dynamic process by coupling in p, essentially as we move from p=2/3 to p=8/9 we need to add enough edges near the origin to be able to take advantage of the high density of edges available far from the origin. The probability of this working given you start from n vertices grows much faster (as n grows) than in the original problem, so you might expect a faster transition.

This is so content-free I’m reluctant even to call it a heuristic. I would be very interested to hear of any more convincing argument for this phenomenon!

REFERENCES

Dekking, Pakes – On family trees and subtrees of simple branching processes (link)