Suppose $A=XDX^{-1}$.
What is the eigenvalue matrix for $A+2I$?
What is the eigenvector matrix?
What is the diagonalisation form of $A+2I=()()()^{-1}$?
Answer:
Here $D$ diagonal matrix of eigenvalues.
If $\lambda_1$ and $\lambda_2$ are the eigenvalues of $A$, then $\lambda_1+2$ and $\lambda_2+2$ are eigenvalues of $A+2I$.
Thus,
$A+2I=X(D+2I)X^{-1}$.
Am I right?
But then what is the eigenvector matrix for $A+2I$?
Help me
You have written down $$A+2I = X(D+2I)X^{-1}$$
That is $$(A+2I)X = X(D+2I)$$
Let $e_i$ be the $i$-th standard unit vector.
$$(A+2I)Xe_i = X(D+2I)e_i$$
$$(A+2I)X_i = X(\lambda_i+2)e_i=(\lambda_i + 2) X_i$$
The columns of $X_i$ are the eigenvectors.
Also, try to prove it directly rather, that is if $Ax=\lambda x$, show that $(A+2I)x=(\lambda + 2) x$.