1. ## 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
pA(t)=(t-1)(t-2)...(t-n).

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

2. ## Re: Polynomial of Matrix

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
pA(t)=(t-1)(t-2)...(t-n).

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

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

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

3. ## Re: Polynomial of Matrix

If the polynomials can be written as

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

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).

4. ## Re: Polynomial of Matrix

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

What is $p(A)$ ? $p(B)$? $p(A)p(B)$? $p(B)p(A)$?
Here is what I came up with...can you let me know if I am right?
$p(A)=a_m A^m + a_{m-1} A^{m-1} +...+ a_0$
$p(B)=a_m B^m + a_{m-1} B^{m-1} +...+ a_0$
$p(A)p(B)=a_m^2 (AB)^m + a_{m-1}^2 (AB)^{m-1} +...+ a_0^2$
$p(B)p(A)=a_m^2 (BA)^m + a_{m-1}^2 (BA)^{m-1} +...+ a_0^2$

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

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