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 $$A = \begin{bmatrix}0 &amp; 1\\0 &amp; 0 \end{bmatrix}$$ 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.
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?
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