Show that matrix $A$ is NOT diagonalizable. Let $A$ be a square matrix $A^2=0$ and $A\neq0$ and show that it is not diagonalizable.
I decided to use the sample matrix of $$A = \begin{bmatrix}0 & 1\\0 & 0 \end{bmatrix}$$ which satisfies the conditions above. 
So my question is: how would I prove this is not diagonalizable. The matrix leads to a eigenvalue of $\lambda=0$ with an algebraic multiplicity of $2$. 
I know that if the algebraic multiplicity and geometric multiplicity are equal, then it is diagonalizable. 
But I am kind of stuck from here since when I use $\det(A-0I)=0$, it just leads to get $x_2=0$ but then also that $x_2=t$ so I don't really know what to do. Any help would be appreciated.
 A: It isn’t enough to prove that your particular sample matrix isn’t diagonalizable: you must show that every non-zero square matrix $A$ such that $A^2=0$ is non-diagonalizable. 
HINT: Suppose that $A^2=0$ and $A$ is diagonalizable. Then there are an invertible matrix $P$ and a diagonal matrix $D$ such that $D=P^{-1}AP$. 


*

*What is $D^2$?  

*What does this tell you about $A$?  

*How does this prove the desired result?

A: The only eigenvalue is 0. If it's diagonalizable, it is similar to the zero matrix, so it's equal to the zero matrix. But it's not.
A: first the eigenvalues of $N = \pmatrix{0&1\\0&0}$ are $0,0$ if $N$ were digonalizable, then the diagonal matrix must be the zero matrix. so $UDU^-1$ will be the zero matrix too. therefore it cannot be $N$ and $N$ is not diagonalizable.  
the same argument holds for any matrix $A \neq 0$ with $A^2 = 0$ reason is $A^2 = 0$ implies $0$ is the only eigenvalue. 
A: If $A^k=0$ (in your case with $k=2$), then any eigenvalue of $A$ satisfies $\lambda^k=0$ (check what happens to an eigenvector) and therefore $\lambda=0$. Then if $A$ is diagonalisable, the only possibility is that $A=0$.
A: You have $det(A^2)=0$ then $det(A)=0$. This implies A has a zero eigenvalues which gives you that A is not diagonalizable
