Vtyjkyuk

If $A in mathbb C^n times n$, $A^2 neq 0$ and $A^3=0$, check if $A$ is diagonalizable.

Now we have $0 lt mboxrank (A) lt n$, what is the next step. Is this about eigenvalues?

Guide:

The eigenvalues are all zero. This matrix has a special name, it is know as a nilpotent matrix.

• Find all eigenvalues of $A$, start from $Ax=lambda x$, try to solve for $lambda$, perhaps by multiplying $A$.




• Now suppose $A$ is diagonalizable, $A=VDV^-1$, observe what is $A$.

$A$ is nilpotent, hence $A$ has only one eigenvalue: $ lambda=0$. Suppose that $A$ is diagonalizable, then there is a basis $z_1,...,z_n$ of $ mathbb C^n$ such that for each $z_j$ we have $Az_j=0$. It follows that $A^2z_j=0$ for $j=1,2,...,n$, hence $A^2=0$ , a contradiction.

Yes, it is about eigenvalues.

If A is diagonalizable, the diagonal matrix have the same eigenvalues than A.

Now, what can you tell about A eigenvalues ?

Hint, if $lambda$ is eigenvalue:

$$exists x, A^3x=lambda^3 x$$

As other people said, you can see that the eigenvalues are zero.
To check the diagonalizable part, remember that the matrix is diagonalizable if and only if the minimal polynomial can be written as a product of first order terms. In your case there is only one eigenvalue and $A^2neq 0$, so the minimal polynomial and the characteristic polynomial must be equal. Do you see what comes next?