How to check whether this matrix is diagonalizable or not. 
Let $\rm A$ be a complex $3\times3$ matrix with $\rm A^3=-1$. Which of the following statements are correct: 
  
  
*
  
*$\rm A$ has three distinct eigenvalues.
  
*$\rm A$ is diagonalizable over $\Bbb C.$
  
*$\rm A$ is triangularizable over $\Bbb C.$
  
*$\rm A$ is non-singular.
  

The characteristic equation for this is $x^3+1=0$.
therefore, $(x+1)(x^2-x+1)=0$ will give you three distinct roots.. so all the answers are correct i think so.. but only $2,3,4$ are correct... why?
 A: If $A^3=-Id$ (the identity matrix), then perhaps $A=-Id$.
A: Recall that a matrix is diagonalizable iff its minimal polynomial
splits into distinct linear factors.
Also recall that the minimal polynomial divides every polynomial $P$
with $P(A)=0$.
In your case you get that $m_{A}|(x-1)(x^{2}-x+1)$.
Now, this does not mean that the roots of $(x-1)(x^{2}-x+1)$ are
the eigenvalues of $A$ but rather that those are the possible values
of the eigenvalues of $A$, for example take 
$$
A=\begin{bmatrix}-1\\
 & -1\\
 &  & -1
\end{bmatrix}
$$
and so the first answer is wrong.
The second answer is correct because of the stated theorem in the
beginning of the answer. 
The third answer is true because every matrix have a Jordan form of
$\mathbb{C}$.
The last answer is correct because $0$ is not a root of $m_{A}$hence
it is not a root of the characteristic polynomial of $A$.
A: For 2), 3) A is diagonalizable Over $\mathbb C$, because A has distinct eigenvalues and every digonalizable matrix is Trianglarizable.
For 4) Eigenvalue of A is non zero, So A is non Singular.
