My memory is fuzzy on the details, but there's a way to show that if you have a series representation of a function that equals that converges to a=that function (i.e., an infinite sums of x^0, x^1, x^2, ..., x^n, ... with n going to infinity that equals f(x) - like the Taylor series of an exponential), that when you apply that to square matrices to represent a matrix function, there's an equivalent representation as a finite matrix polynomial. You can take the highest power of the equivalent finite polynomial as low as one less than the order of the matrix (so a matrix function of NxN matrix can be written as a N-1 degree matrix polynomial).

In other words, for a square NxN matrix A and functions satisfying the appropriate conditions, this convergent infinite sum of increasing powers of A:

c0 + c1A + c2A^2 + ... + cnA^n + ...

can be written as the n-1 degree matrix polynomial:

b0 + b1A + b2A^2 + .... + b_{n-1} A^(n-1)

The coefficients b0, ..., b_{n-1} are different than the coefficients c0, c1, ... for the infinite series, but they evaluate to the same thing.

Critical to the point you're making, the b0, b1, ..., b_{n-1} are different for differing choices of A, because the bn coefficients depend on the eigenvalues of the matrix. So you really have b0(A) + b1(A) * A, etc. The coefficients of the infinite series (c0, c1, ...) do not have this dependence on A.

That makes the 1x1 case tautological, because

b0(A) = f(A)

That is, the finite series representation is the value of the function at its argument.

There's some info on some of that here:

https://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theore...

Ok, now I see what you are saying. Thanks!

From Wikipedia[1]:

When X is an n×n diagonal matrix then exp(X) will be an n×n diagonal matrix with each diagonal element equal to the ordinary exponential applied to the corresponding diagonal element of X.

Perhaps that's what GP meant?

[1]: https://en.wikipedia.org/wiki/Matrix_exponential