finding the Jordan canonical form

Question

I have a question about the Jordan normal form:

Assuming we have a $3$-dimensional vector Space $V$ over the field of complex numbers. We have a linear map $f: V \to V$ and we know that the rank of the image of $f$ is $2$. Furthermore we know that $f$ has no non-zero eigenvalues.

Apparently it follows that the Jordan normal form must be $$ \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}. $$

My question is why does the Joran normal form look like this?

Answer 1

Because all eigenvalues of $f$ are zero, the Jordan form (up to rearranging the blocks) must be one of the following: $$ \pmatrix{0&0&0\\0&0&0\\0&0&0}, \quad \pmatrix{0&1&0\\0&0&0\\0&0&0}, \quad \pmatrix{0&1&0\\0&0&1\\0&0&0}. $$ Now, which of these has a rank of $2$?