Commuting matrices

As an application for diagonalization and the diagonal form, consider the question of discovering which matrices $N$ commute with a given matrix $M$:

$$\begin{eqnarray*} M N & = & N M. \end{eqnarray*}$$

Choosing the preferred basis,

$$\begin{eqnarray*} V M U V N U & = & V N U V M U \\ \Lambda K & = & K \Lambda, \end{eqnarray*}$$

after introducing $K = V N U$. In more explicit detail,

$$\begin{eqnarray*} \left[ \begin{array}{llll} \lambda_1 & . & . & . \\ . & \... ... . & \lambda_2 & . & . \\ . & . & . & . \end{array} \right] \end{eqnarray*}$$

in general summarized by the relations involving the individual matrix elements,

$$\begin{eqnarray*} (\lambda_i - \lambda_j) k_{ij} & = & 0. \end{eqnarray*}$$

Once again, distinct eigenvalues yield a clearcut conclusion, whereas multiple eigenvalues require further analysis. The conclusion is: $K$ is diagonal too, but in the presence of multiplicity, some further adjustment, resolved by a modified basis, may be necessary to manifest both matrices in that form.

The conclusion which we seem to have is that matrices commute only when one is a function of the other, or better yet, are common functions of a third matrix which has distinct eigenvalues. Which function, exacty? It is a polynomial whose degree is no higher than the dimension of the matrix, but to find out which one, the eigenvalues with which the Lagrange interpolation polynomials operate have to be taken over to the maclaurin basis to discern the coefficients of the polynomial.

The result also casts some doubt upon whether we know how to find the square root of a matrix; so far we really only know how to get square roots which commute with the matrix; there may still be others.

Another curiosity: the exponential of a matrix is just a polynomial, even though the exponential is an infinite series; the same observation holds for sines, cosines, and whatever other function. The infinite series still plays a role; it is one way to calculate the appropriate function of the eigenvalues before inverting the lagrange interpolation to get the polynomial.