$A\in R^{n\times n}$ is diagonalizable. Prove that $A^2+3I$ is invertible

Question

So I'm having a lot of trouble with the proof. I see that $A^2+3I$ is diagonalizable, but not how to move on to it being invertible. Any direction would be appreciated!

Answer 1

Depends upon your meaning for "diagonalizable" $$ A = \left( \begin{array}{cc} 1 & -2 \\ 2 & -1 \end{array} \right) $$ This one diagonalizes as $$ \left( \begin{array}{cc} i \sqrt 3 & 0 \\ 0 & - i \sqrt 3 \end{array} \right) $$ and $$ A^2 + 3 I = 0 $$

Answer 2

If $A$ is diagonal, then $A^2+3I$ is also diagonal. Invertibility of $A^2+3I$ depends on that the diagonal entries are nonzero, and...

If $A=P^{-1}DP$ for a diagonal matrix $D$, then $A^2+3I=P^{-1}(D^2+3I)P$, and so...