# But will they commute?

While working on a project, I had to learn a couple on stuff on semigroups of bounded operators, a really useful tool if, for example, you have some kind of Cauchy problem of the following form,

$\left\{\begin{array}{lr}\frac{du(t)}{dt}=Au\\u(0)=x\end{array}\right.$

where A is an operator.

Reading about semi-groups reminded me of another problem that I had seen as an undergrad, the problem of commutativity of matrix multiplication. Well, it’s not a really a problem, more of a consequence of the way matrix multiplication is defined. But let’s take it from the top.

Let’s say that we have two square matrices A and B of dimensions $2\times 2$. Obviously there are many ways in which we could define a multiplication between them. For example, we might want to multiply them element by element and get back a new matrix $C=\left (a_{i,j}b_{i,j}\right)_{1\leq i,j\leq 2}$. So then, why is matrix multiplication  that “complicated”?

The reason is this. A matrix represents a linear mapping between two vector spaces and since it’s useful to consider compositions of those linear transformations, the matrix product is just that. In other words, if A represents the linear mapping T and B represents the linear mapping F, then AB represents the composition T(F).

Since matrix multiplication is defined in this way, it’s not a surprise that generally $AB\neq BA$, since we can write down linear transformations T and F for which the different compositions T(F) and F(T) will be unequal. Take for example $Tx=2x$ and $Fx=x/2+1$, then

$T(F(x))=x+2\neq F(T(x))=x+1$

So a good question is, if I have a matrix A, can I somehow talk about the structure of a B such that $AB=BA$ or, in more modern notation, such that the commutator $[A,B]=AB-BA=0$? Since we can represent linear mappings as matrices, this question extends naturally to them as well.

Extra notation; let’s denote the set of all B such that $[A,B]=0$ by $Com(A)$.

Let’s take a look at a special case.

Assumptions : Let A be square & diagonalizable.

So, we have a full set of eigenvectors, written as columns in a matrix V and a set of eigenvalues $\lambda_{1},\ \lambda_{2},\ldots$, which we write as a diagonal matrix D. Then we can write A in the form, $A=VDV^{-1}$.  Supposing that A is $2\times 2$, something that we can do without loss of generality as far as the dimensions are concerned, that means that :

$A=VDV^{-1}=V\left(\begin{array}{lr}\lambda_{1} &0\\ 0& \lambda_{2}\end{array}\right)V^{-1}$

Can we now find a matrix $B\in Com(A)$?

The answer is yes. Just suppose that B has the same eigenvectors as A but different eigenvalues. Then

$B=V\left(\begin{array}{lr}\gamma_{1}&0\\ 0&\gamma_{2}\end{array}\right)V^{-1}$

Then, we can see that $AB=BA$ or, in other words, $B\in Com(A)$.

I wonder if there is a way to find non-trivial examples of non-diagonalizable matrices in Com(A) (Jordan normal form, perhaps?).  Any thought on that would be welcome. 😀

1. Klaus says:

Well, commuting with a given matrix A boils down to a system of linear equations in the coefficients of B, making the solution straightforward if the dimension is not too large.

1. Konstantinos says:

Well yes, of course, this should be always possible if the matrix A is invertible. Thanks for that!

I was hoping for something that would give me a hint on some kind of structure of Com(A). For example, from the previous eigenvalue – eigenvector idea, I can see that [AB]=0 if and only if A, B preserve the direction of the same (eigen)vectors in $\mathbb{R}^n$.

But of course, noone has high hopes for special structure in a general case (like when having a random A). 🙂