$A \in M_{3}(\Bbb{R})$, then diagonalizability of $A$ in the following two cases 1) $(A-3I)^2 = 0$ and 2) $A^2 + I =0$

Question

Let $A$ be a $3 \times 3 $ matrix , $A \in M_{3}(\Bbb{R})$; then I was thinking of diagonalizability of $A$ in the following two cases 1) $(A-3I)^2 = 0$ and 2) $A^2 + I =0$.

In the first problem 1) I think Since $(A- 3I)^2$ matrix has all the eigenvalues zero so $A = 3I$ which is a diagonal matrix and hence diagonalizable.

In the 2nd problem, I think since $A^2 + I = 0$ so the two eigenvalues of $A$ are $i , -i$ but I was thinking what the next eigenvalue would be? if all the eigenvalues would have been distinct then I could say that the matrix is diagonalizable.

Are the above arguments correct?

Answer 1

Hint:

1. There are $3 \times 3$ matrices $B\ne0$ such that $B^2=0$.

2. If $A^2=-I$, what can you say about $\det A$?

Answer 2

A matrix is certainly diagonalisable over an algebraically closed field if $f(A)=0$ where $f$ is a polynomial with no repeated zeros. Over $\Bbb C$, $x^2+1=(x-i)(x+i)$.

Here is a $2$-by-$2$ matrix with $B^2=0$: $B=\pmatrix{0&1\\0&0}$. If there was a similar $3$-by-$3$ example and $A=B+3I$, what then?