Eigenvalues, polynomials and minimal polynomials

Question

I have proved (a) by:

Let $\lambda$ be an eigenvalue of $AB$

$ABv=\lambda*v$

Then $BABv=\lambda*B*v$ so Bv is an eigenvector of BA with eigenvalue $\lambda$.

For B, I have found the formula in my textbook, however, cannot see a proof, it is stated as 'elementary'. Also, having trouble with (c), so would appreciate any help.

Answer 1

Suppose

$$p(x)=\sum_{k=0}^n a_kx^k\implies P(AB)\cdot A=\sum_{k=0}^n a_k(AB)^k\cdot A\;\;(*)$$

and now observe that using associativity of matrix product in each monomial, we get

$$(AB)^kA=(AB)(AB)(AB)\cdot\ldots\cdot(AB)\cdot A= A\cdot (BA)(BA)\cdot\ldots\cdot (BA)$$

so that we get

$$(*)=A\sum_{k=0}^n a_k(BA)^k=A\cdot p(BA)$$

Indeed, elementary...but not straigthforward trivial.

Check how this can help you, if at all, with (c), too.