Let $S$ be a diagonalizable matrix and $S+5T=I$. Then prove that $T$ is also diagonalizable. 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.
 A: This proof is great!
Here's another way.
Recall that a linear transformation $F:V\to V$ is diagonalizable if and only if there exists a basis of $V$ consisting of eigenvectors of $F$.
In your case, we have two linear transformations $S,T:V\to V$ such that 
$$
T=\alpha(I-S)
$$
where $\alpha=1/5$. Furthermore, $S$ is diagonalizable so there exists a basis $\{s_1,\dotsc,s_n\}$ of $V$ consisting of eigenvectors of $S$. If $\{\lambda_1,\dotsc,\lambda_n\}$ are the corresponding eigenvalues of $S$, then
$$
Ts_k=\alpha(I-S)s_k=\alpha(s_k-Ss_k)=\alpha(s_k-\lambda_ks_k)=\alpha(1-\lambda_k)s_k
$$
This proves that each $s_k$ is also an eigenvector of $T$ with corresponding eigenvalue $\alpha(1-\lambda_k)$. That is, $\{s_1,\dotsc,s_n\}$ is a basis of $V$ consisting of eigenvectors of $T$. Hence $T$ is diagonalizable.
A: 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 $F$.
A: If a matrix $A$ is diagonalizable, then $\alpha A$ is, for every scalar $\alpha$ (easy). Thus you just need to show that $5T$ is diagonalizable. Since
$$
P(5T)P^{-1}=P(I-S)P^{-1}=I-PSP^{-1}=I-D
$$
we have that $5T$ is similar to a diagonal matrix.
