Suppose $T^2=0$. Prove that the Jordan canonical form of $T$ consists of $\dim(\ker T)$ Jordan blocks, $\dim(Im T)$ of which are $2\times 2$ blocks. Question: Let $T:V\rightarrow V$ be a linear transformation satisfying $T^2=0$.  Prove that the Jordan canonical form of $T$ consists of $\dim(\ker T)$ Jordan blocks, $\dim(Im T)$ of which are $2\times 2$ blocks.
Thoughts: Since $T^2=0$, the min poly is either $x$ or $x^2$, so I should consider cases here.  If the min poly is just $x$, could I claim that $T=0$ then?  Since the min poly is either $x$ or $x^2$, could I say that the only eigenvalue of $T$ is zero?  So I could then play with some linearly independent stuff... but I am not seeing how everything is coming together.  Any help is greatly appreciated!  Thank you.
 A: 
If the minimal polynomial is $x$, could I claim that $T=0$ then?

Yes; this follows by the definition of the minimal polynomial.


If the minimal polynomial is either $x$ or $x^2$, could I say that the only eigenvalue of $T$ is zero?

Yes. Consider a Jordan block $B$ for a nonzero eigenvalue. Can $B^k$ ever be the zero matrix for some $k \ge 1$?

If $T^2=0$, the Jordan blocks must have size $2\times 2$ or smaller. (If there is a larger Jordan block $B$, then $B^2 \ne 0$.) You also know the eigenvalues are all zero. So all that remains is to understand how many $2 \times 2$ blocks there are and how many $1 \times 1$ blocks there are.
Play around with such matrices to understand why the number of blocks is equal the the dimension of the kernel. The rank-nullity theorem will then imply the dimension of the image is the number of $2 \times 2$ blocks.
The column space is spanned by the nonzero columns of the matrix. Each $2 \times 2$ block contributes one nonzero column, and these columns are linearly independent, so the dimension of the image is the number of $2 \times 2$ blocks. Then you can use the rank-nullity theorem to conclude that the dimension of the kernel is $\dim(V)-(\text{number of $2 \times 2$ blocks}) = \text{number of blocks}$.
