# Thread: Matrix Formula

1. ## Matrix Formula

I have this:
A=any square matrix over any field F
p(X)=any polynomial over F
P=any invertible matrix over F
B=inverse(P)AP=P^(-1)AP

I need to show that p(B)=inverse(P)p(A)P=P^(-1)p(A)P.
How would I prove this?

2. No it looks like what I wrote.

3. Is there anyway to show it more algebraically?

He gives a hint: First do this in the case p(X)=X^m.

4. Sorry about that.
It used the important property of similar matricies.
$(P^{-1}AP)^n=P^{-1}A^nP$.
Let the polynomial F[x] be,
$f(x)=a_0+a_1x+a_2x^2+...+a_nx^n$.
Then,
$f(B)=a_0I+a_1B+a_2B^2+...+a_nB^n$
And,
$f(P^{-1}AP)=a_0I+a_1(P^{-1}AP)+a_2(P^{-1}AP)^2+...+a_n(P^{-1}AP)^n$
Use the property mentioned about,
$f(P^{-1}AP)=a_0(P^{-1}P)+a_1(P^{-1}AP)+a_2(P^{-1}A^2P)+...+a_n(P^{-1}A^nP)$
You can use left-right distributive laws in a matrix,
$f(P^{-1}AP)=P^{-1}(a_0+a_1A+a_2A^2+...+a_nA^n)P=P^{-1}f(A)P$
But,
$f(B)=f(P^{-1}AP)$
Thus,
$f(B)=P^{-1}f(A)P$

(The only mistake with you question is that you said a polynomial over a field. That is not what I assumed).