TITLE: Let $S$ be a diagonalizable matrix and $S+5T=I$. Then prove that $T$ is also diagonalizable.
QUESTION [13 upvotes]: My solution:

Since $S$ is diagonalizable, so we can write $S=P^{-1}DP$, where $P$ is an invertible matrix and $D$ is a diagonal matrix.
Now $5T=I-S=P^{-1}P-P^{-1}DP=P^{-1}(I-D)P$. So $T=P^{-1}\frac{1}{5}(I-D)P$.     Since $I-D$ is also a diagonal matrix, hence $T$ is diagonalizable.

Is my proof correct? Can it be done in another way? thanks.

REPLY [2 votes]: Proposition 1. If $F$ is any field and $A\in M_{n}(F)$ is a diagonalizable matrix then, for every polynomial $f(x)$, $f(A)\in M_{n}(F)$ is also diagonalizable.
Proof. We have $A=P^{-1}DP$ for some invertible matrix $P$ and diagonal matrix $D$. Then
$f(A)=P^{-1}f(D)P$. Since $f(D)$ is diagonal, $f(A)$ is diagonalizable.
Corollary. If the minimal polynomial of a matrix $A\in M_{n}(F)$ is a product of distinct linear factors then, for every polynomial $f(x)$, the minimal polynomial of $f(A)$ is also a product of distinct linear factors. 
Proposition 2. If $F$ is any field and $A\in M_{n}(F)$ is a triangulable matrix then, for every polynomial $f(x)$, $f(A)\in M_{n}(F)$ is also triangulable.
Proof. We have $A=P^{-1}\Delta P$ for some invertible matrix $P$ and (upper or lower) triangular matrix $\Delta$. Then
$f(A)=P^{-1}f(\Delta)P$. Since $f(\Delta)$ is (upper or lower) triangular, $f(A)$ is triangulable.
Corollary. If the minimal polynomial of a matrix $A\in M_{n}(F)$ factors completely over $F$ then, for every polynomial $f(x)$, the minimal polynomial of $f(A)$ also factors completely over $F$.