Q:

Matrix of T, a linear transformation when Im T = Ker T

Let $V$ be a finite dimension vector space, $T:\ V \to V$ a linear transformation, and assume that $\operatorname{\rm Ker} T = \operatorname{\rm Im} T$. Prove that there is a basis $B$ of $V$, so that the matrix representation of $T$ in that basis is
$$\begin{bmatrix} 0 & I \\ 0 & 0 \end{bmatrix},$$
where every block of the matrix is of order $n/2$.
I already know that the dimension of $V$ is even, and didn't succeed to find which basis is it. I appreciate your help.

A:

Choose an arbitrary basis $(b_1,..b_n)$ of $\ker T$ and extend it to a basis $(b_1,..,b_n,c_1,..c_n)$ of the whole space $V$, which has dimension $2n=\dim\ker T+\dim{\rm im\,}T$ by the dimension theorem.
Then forget $b_i$, and in place of them consider $d_i:=Tc_i$.
Now conclude that $(d_1,..,d_n)$ forms a basis of ${\rm im\,}T$, and that the matrix of $T$ will have the required form in the basis $(d_1,..,d_n,c_1,..,c_n)$.

