Prove that if matrix $A$ is diagonalizable then for all $c\in \mathbb R$ $cI-A$ is diagonalizable 
Prove that if matrix $A$ is diagonalizable then for all $c\in \mathbb R$ $cI-A$ is diagonalizable.

Can this be proven in this way as follows?
$A$ is diagonalizable then an eigenvalue $c\in \mathbb R$ exists such that $Av=cv$.
Then:
$$
Av=cv \Leftrightarrow Av-cv=0\\
(A-cI)v=0 | \cdot (-1)\\
(cI-A)v=0\\
(cI-A)v=c\cdot0
$$
Therefore $c$ is an eigenvalue of $(cI-A)$.
Please provide feedback to my question first and foremost (that is whether this proof works).
 A: The proof you have does not show that $c$ is an eigenvalue of $cI-A$. In fact, we cannot say that if $c$ is an eigenvalue of $A$ then $c$ is an eigenvalue of $cI-A$ in general.
What we can do: since $A$ is diagonalizable, we have $A=P^{-1}DP$ for some invertible matrix $P$ and some diagonal matrix $D$ (of course all these matrices are square and of the same size). Hence, we have:
$$cI-A=cI-P^{-1}DP=cP^{-1}P-P^{-1}DP=P^{-1}(cI-D)P$$
Therefore, $cI-A$ is diagonalizable since $cI-D$ is a diagonal matrix.

To address Widawensen's comment, suppose that $A$ is diagonal and therefore $A=P^{-1}DP$ as before. Then, for $a_i\in\Bbb R~~~i\in\{0,1,\ldots,n\}$ we have:
$$\begin{align}&a_nA^n+a_{n-1}A^{n-1}+\cdots+a_1A+a_0I\\=&a_n(P^{-1}DP)^n+a_{n-1}(P^{-1}DP)^{n-1}+\cdots+a_1(P^{-1}DP)+a_0P^{-1}P\\=&a_nP^{-1}D^nP+a_{n-1}P^{-1}D^{n-1}P+\cdots+a_1P^{-1}DP+a_0P^{-1}P\\=&P^{-1}(a_nD^n+a_{n-1}D^{n-1}+\cdots+a_1D+a_0I)P\end{align}$$
Note that we used the fact that $(P^{-1}DP)^k=P^{-1}D^kP~~~\forall k\in\Bbb N$. Therefore, $a_nA^n+a_{n-1}A^{n-1}+\cdots+a_1A+a_0I$ is diagonalizable since $a_nD^n+a_{n-1}D^{n-1}+\cdots+a_1D+a_0I$ is diagonal (because $D^k$ is diagonal $\forall k\in\Bbb N$).
A: Nope, you are working with only at most $n$ particular values of $c$,which are just the eigenvalues.,  hence no matter how hard you  work, you are still not addressing the quesiton.  You should work with any $c\in \mathbb{R}$.
Also, the right conclusion for your working is $0$ is an eigenvalue for $cI-A$. Your conclusion can also be extended to if $v$ is an eigenvector of $A$ with eigenvalue $c$, then $v$ is also an eigenvector of $cI-A$ with eigenvalue $0$. You have not address whether we have sufficient eigenvectors corresponding to other eigenvalues. 
Hint: Write $A$ as $PDP^{-1}$ and express $I$ in terms of $P$.
