Let $\mathbb{K}$ be a field and $f \in \mathbb{K}[X]$ a polynomial over $\mathbb{K}$. Let $A \in \mathbb{K}^{n \times n}$ be a diagonalizable matrix. Show that $f(A)$ is also diagonalizable.
I hope you can help me, I have problems with showing something in general without examples.
On
A generic polynomial over $\mathbb{K}$ has the form $p(x) = c_0 + c_1x + c_2x^2 + ... + c_nx^n, c_i \in \mathbb{K}$.
When you use a matrix $A$ as input for the polynomial, you get $ f(A) = c_0I + c_1A + c_2A^2 + ... + c_nA^n$.
Now you just have to conclude that, since $A$ is diagonalizable, all of the matrices in the above polynomial are also diagonalizable. If you understand diagonalizability, this is rather obvious. But I'll just say that every matrix in the above polynomial satisfies
$$ PA^nP^{-1} = D^n$$
Since $A$ is diagonalizable, there exists a diagonal matrix $D$ and an invertible matrix $S$ such that $D=SAS^{-1}$. If we write out $f(x)=c_nx^n+...+c_1x+c_0$, then what happens if we multiply the matrix $f(A)$ on the left by $S$ and on the right by $S^{-1}$? You might find the fact that $(SAS^{-1})^k=SA^kS^{-1}$ to be helpful.