Matrix diagonalizable or not Let $A$ is in $M_3(\mathbb R^3)$ which is not a diagonal matrix. Pick out the cases when $A $ is diagonalizable over $\mathbb R$:
a. when $A^2=A$;
b. when $(A-3I)^2=0$;
c. when $A^2+I=0$.
My attempt is if $A$ is diagonalizable then there is some invertible $P$ s.t. $PAP^{-1}$=$D$. 
Then case a. gives $D^2$=$D$. then $D$ is either $0$ or $I$. Which gives $A$=$0$ or $I$..a contradiction to the fact $A$ is not diagonal. But I am not sure about my approach. Similarly I arrive contradiction for other cases. Please help.
 A: Result :
A matrix is diagonalizable if its minimal polynomial has no repeated roots.
For first case $A^2=A$ minimal polynomial could be :


*

*$x$ What does this mean if $x$ is minimal polynomial for $A$.. does something goes wrong?

*$x-1$ What does this mean if $x$ is minimal polynomial for $A$.. does something goes wrong?

*and the other possibility is ???


You should be able to complete that i believe.
For second case $(A-3I)^2=0$ what could be the minimal polynomials?
For third case $A^2+I=0$ minimal polynomial would be $???$
That polynomial do not have roots in $\mathbb{R}$ so there is no question of diagonalizability over $\mathbb{R}$.
A: Here's how I would dispose of these issues:
For (c.), note that a matrix $A \in M_3(R)$ has at least one real eigenvalue, since its characteristic polynomial is of degree 3, and real polynomials of odd degree have at least one real root.  But if $A^2 + I = 0$, the eigenvalues $\lambda$ of $A$ must satisfy $\lambda^2 + 1 = 0$, i.e. $\lambda = \pm i$.  This shows there is no matrix $A \in M_3(R)$ satisfying $A^2 + I = 0$.  Thus the proposition, 
"$A \in M_3(R) \; \text{and} \; A^2 + I = 0 \Rightarrow A \; \text{is diagonalizable over} \; R.$"
is vacuously true.
For (b.), let
$A = \begin{bmatrix} 3 & 1 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix}; \tag{1}$
then
$(A - 3I)^2 = N^2 = 0, \tag{2}$
where
$N = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}. \tag{3}$
We now see that $A$ cannot be diagonalizable over $R$; for, since the only eigenvalue of $A$ is $3$ by virtue of $(A - 3I)^2 = 0$, if there were a nonsingular real matrix $P$ with
$PAP^{-1}$ diagonal, we would have
$PAP^{-1} = 3I; \tag{4}$
but (4) implies $A = P^{-1}(3I)P = 3I$, contradicting (1).  This example can be generalized in the following sense:  any non-diagonal $A$ satisfying $(A - 3I)^2 = 0$ may clearly be written as $A = 3I + N$, with $N \ne 0$ satisfying $N^2 = 0$.  The only eigenvalue of such $A$ is $3$; thus $PAP^{-1}$ diagonal forces $A = 3I$, exactly as we have just seen.  This contradicts $N \ne 0$.  
Finally, for (a.), use the Jordan form: if $A$ were not diagonalizable, each Jordan block J would have to satisfy $J^2 = J$; but $J = \lambda I + N$ with $N = [n_{ij}]$, and $n_{ij}  = 1$ iff $j = i+ 1$, $n_{ij} = 0$ otherwise; so $J^2 = \lambda^2 I + 2\lambda N + N^2 \ne J$; thus $A$ can be diagonalized.
Hope this helps.  Cheers,
and as always, Fiat Lux!!!
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