# Polynomial of Matrix

• Nov 26th 2011, 03:16 PM
page929
Polynomial of Matrix
Let A,B belong to Mn and AB=BA. Let p(•) be any polynomial, say of degree m. Show that p(A)p(B) = p(B)p(A).

The only thing that I have in my notes states that

I have no idea even how to begin this. Any help would be great. Thanks!
• Nov 26th 2011, 03:28 PM
Also sprach Zarathustra
Re: Polynomial of Matrix
Quote:

Originally Posted by page929
Let A,B belong to Mn and AB=BA. Let p(•) be any polynomial, say of degree m. Show that p(A)p(B) = p(B)p(A).

The only thing that I have in my notes states that

I have no idea even how to begin this. Any help would be great. Thanks!

p(•)=$\displaystyle p(x)=a_m x^m + a_{m-1} x^{m-1} + . . . +a_0$

What is $\displaystyle p(A)$ ?$\displaystyle p(B)$? $\displaystyle p(A)p(B)$? $\displaystyle p(B)p(A)$?
• Nov 26th 2011, 03:49 PM
uasac
Re: Polynomial of Matrix
If the polynomials can be written as

$\displaystyle p(A)=\sum_{j=0}^ma_jA^j$

Then I would use the discrete convolution formula and try to figure out the math. Be careful with the indices, it seems to me you have to play a little bit with those (I haven't done the computations but this seems the way to go).
• Nov 27th 2011, 06:42 AM
page929
Re: Polynomial of Matrix
Quote:

Originally Posted by Also sprach Zarathustra
p(•)=$\displaystyle p(x)=a_m x^m + a_{m-1} x^{m-1} + . . . +a_0$

What is $\displaystyle p(A)$ ?$\displaystyle p(B)$? $\displaystyle p(A)p(B)$? $\displaystyle p(B)p(A)$?

Here is what I came up with...can you let me know if I am right?
$\displaystyle p(A)=a_m A^m + a_{m-1} A^{m-1} +...+ a_0$
$\displaystyle p(B)=a_m B^m + a_{m-1} B^{m-1} +...+ a_0$
$\displaystyle p(A)p(B)=a_m^2 (AB)^m + a_{m-1}^2 (AB)^{m-1} +...+ a_0^2$
$\displaystyle p(B)p(A)=a_m^2 (BA)^m + a_{m-1}^2 (BA)^{m-1} +...+ a_0^2$
• Nov 27th 2011, 07:20 AM
Deveno
Re: Polynomial of Matrix
actually, what you have written down is p(A), p(B), p(AB), and p(BA).

p(A)p(B) is the product of two polynomials, and will be of degree 2m. and not every coefficient will be a perfect square.

for example, if p(x) was x^2 + 1, p(A)p(B) = (A^2 + I)(B^2 + I) = A^2B^2 + A^2 + B^2 + I = (AB)^2 + A^2 + B^2 + I, not (AB)^2 + I.

the formula is messy, because you have for each term of degree k, several things to sum together. for example,

the next term after $\displaystyle a_m^2A^mB^m$ wil be $\displaystyle a_ma_{m-1}(A^mB^{m-1} + A^{m-1}B^m)$, and the term after that is:

$\displaystyle a_ma_{m-2}A^mB^{m-2} + a_{m-1}^2A^{m-1}B^{m-1} + a_ma_{m-2}A^{m-2}B^m$.