How can one find the possible Jordan forms for a $2 \times 2$ matrix $A$ satisfying $A^2 = 0$? I need to find all the possible Jordan normal forms for a $2 \times 2$ matrix $A$ satisfying $A^2 = 0$.  
If I have any $2 \times 2$ matrix, I can have either have one eigenvalue of multiplicity $2$ or two different eigenvalues of multiplicity $1$.  I don't see how $A^2 = 0$ would make it any different?  Which possibility can I eliminate and why?
 A: The Jordan normal form has blocks of the shape $$\left[\begin{array}{cccc}
\lambda&1&0&0\\
0&\ddots&\ddots&0\\
0&0&\lambda&1\\
0&0&0&\lambda
\end{array}\right]$$
For 2x2 matrix you have 2 possibilities: 


*

*2 blocks of $1 \times 1$ each which would just be $[\lambda]$. 

*1 block of $2\times 2$ which would be $\left[\begin{array}{cc}
\lambda&1\\
0&\lambda\\
\end{array}\right]$


Now maybe it gets easier.
A: $A^2 = 0$ tells you that your characteristic polynomial is $x^2$. Since the minimal polynomial divides the characteristic polynomial, this tells you that your minimal polynomial has to be either $x$ or $x^2$.
Case 1: Min poly = $x$.
In this case we have that $A = 0$ and is already in Jordan Form.
Case 2: Min poly = $x^2$.
In this case the Jordan Form is $$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}_.$$
A: One can take advantage of the small size of the matrix in question.
The condition $A^2=0$ tells you that the only possible eigenvalue is $\lambda=0$. Thus the only possible Jordan forms for $A$ are
$$
\begin{pmatrix}
0&0\\
0&0
\end{pmatrix},\quad
\begin{pmatrix}
0&1\\
0&0
\end{pmatrix}.
$$
Now one can check immediately that any matrix $B$ similar to any one of them satisfies $B^2=0$. 

If you are considering larger size matrices, the general technique would be the minimal polynomials. 
A: Let J be the Jordan canonical form for A. Then there exists an invertible matrix M such that

A = M-1* J * M

Then

A2 = (M-1* J * M) * (M-1* J * M) = M-1* J2 * M = 0

Suppose there is some nonzero eigenvalue λ of A. Then either λ is the only eigenvalue or there is some other (possibly zero) eigenvalue λo.

Case 1: J = $\begin{bmatrix}λ & 1\\0 & λ \end{bmatrix}$

J2 = $\begin{bmatrix}λ^2  & 2λ\\0 & λ^2 \end{bmatrix}$ = 0
==> λ = 0, which is a contradiction.

Case 2: J = $\begin{bmatrix}λ & 0\\0 & λo \end{bmatrix}$

J2 = $\begin{bmatrix}λ^2 & 0\\0 & λo^2 \end{bmatrix}$ = 0
==> λ = 0, which is a contradiction.


So there can be no non-zero eigenvalue of A. Therefore the only two possible Jordan canonical forms for A are
J = $\begin{bmatrix}0 & 1\\0 & 0 \end{bmatrix}$ or J = 0.
In the latter case, J = 0 implies A = M-1 * J * M = 0. Note that in general, for an nxn matrix A such that An = 0, the Jordan canonical form of A must be composed solely of Jordan blocks with eigenvalue 0. By the same reasoning above, An = 0 implies Jn = 0. If J is represented in block diagonal form, then we see that every block raised to the nth power is the zero matrix. Suppose we have some block Jλ of size greater than one. Then Jλn will have λn along the diagonals, so Jλn = 0 implies λ = 0.
