I've been trying to solve this issue:
Let $A$ be a diagnosable $n\times n$ matrix. It's characteristic polynomial is: $p(t) = a_n t^n +a_{n-1}t^{n-1}+…+a_1t+a_0$.
Given the square $n\times n$ matrix $p(A)=a_nA^n+a_{n-1}A^{n-1}+…+a_1A+a_0$, prove that $p(A)=0$
We can deduce that $a_n = 1$, and that $a_{n-1}=-tr(A)$, or the minus the sum of all eigenvalues.
But I'm not sure what more can we find out about the given characteristic polynomial.
Any help would be appreciated.
EDIT: Thanks to the help of Sam Freedman, a solution was found.
Matrix A is diagnosable, which means it is similar to a diagonal Matrix D:
$A = P^-1DP$
That means we can define $p(A)$ with matrix D:
$p(P^-1DP) = P^-1(D^n +a_{n-1}D^{n-1}+…+a_1D+a_0)P$
We can conclude that matrix $p(A)$ is similar to a diagonal matrix, described above.
Matrix $D$ can be broken down to: $diag(d_1,…,d_n)$. And the diagonal matrix $p(A)$ is similar to can will have the value: $d_i^n+a_{n-1}d_i^{n-1}+…+a_1d_i +a_0$, which means those are the eigenvalues of $p(A)$.
Now, thanks to knowing the characteristic polynomial of $A$, we know that the group $d_1,…,d_n$ are the roots of said polynomial, meaning they send it to 0.
And the form of each eigenvalue of $p(A)$ is equal to the form of said characteristic polynomial. Which means each eigenvalue of $p(A)$ is equal to 0.
The diagonal matrix $p(A)$ is similar to must be the zero matrix. And the zero matrix is the only matrix which is similar to itself, and therefore we can conclude and claim:
$p(A) = 0$
Many thanks, Sam Freedman!
Best Answer
This is the famous Cayley-Hamilton theorem, but we can also prove this special diagonalizable case directly.
Writing $A = P^{-1} D P$, we use a standard trick to compute out \begin{align*} p(A) = p( P^{-1} D P) &= ( P^{-1} D P)^n + a_{n - 1}( P^{-1} D P)^{n - 1} + \dots + a_1 ( P^{-1} D P) + a_0I \\ &= P^{-1} D^n P + a_{n - 1}P^{-1} D^{n - 1} P + \dots + a_1 ( P^{-1} D P) + a_0I\\ &= P^{-1}( D^n + a_{n - 1} D^{n - 1} + \dots + a_1 D + a_0I ) P. \end{align*}
Now we've reduced the problem to the case of a diagonal matrix. Given what you know about the characteristic polynomial of a diagonal matrix, can you go from here?