$\ T: V \rightarrow V $ where $\ V $ is a space of dimension $n$ .also $\ker T = \operatorname{Im}T $ and I need to prove there is a basis $\ B$ of $V$ such that $[T]_B$ will be : $\begin{bmatrix} 0 & I\\ 0 & 0 \end{bmatrix}$ (four blocks, each is $\frac{n}{2} \times \frac{n}{2}) $
Now I understand that if $\ker T = \operatorname{Im}T $ then $\dim(V) = \dim(\ker T) + \dim(\operatorname{Im}T) $ which means $\ n $ is even number. But also if $\ker T = \operatorname{im}T $ then for every $v \in V, T(v) \in \ker T $ because $\operatorname{Im}T \subseteq \ker T$ so wouldn't the representing matrix will all be just zeros?
KetT=ImgT therefore dim Img T + dim Img T = dim Img T + dim Ker T = dim V therefore n is even KetT=ImgT therefore for each v in V T(T(v))=0.
let (v(1),...,v(n/2)) be a basis of ImgT Then for each i between 1 and n/2 there exists a u(i) in V such that T(u(i))=v(i) (by the definition of Img)
let a(1)u(1)+a(2)u(2)+...+a(n/2)u(n/2)+b(1)v(1)+b(2)v(2)+...+b(n/2)v(n/2)=0 be a linear combination. then T(a(1)u(1)+a(2)u(2)+...+a(n/2)u(n/2)+b(1)v(1)+b(2)v(2)+...+b(n/2)v(n/2))=0. then T(a(1)u(1)+a(2)u(2)+...+a(n/2)u(n/2))+T(b(1)v(1)+b(2)v(2)+...+b(n/2)v(n/2))=0 then T(a(1)u(1)+a(2)u(2)+...+a(n/2)u(n/2))=0 then a(1)T(u(1))+a(2)T(u(2))+...+a(n/2)T(u(n/2)) = 0 then a(1)v(1)+a(2)v(2)+...+a(n/2)v(v/2) = 0 because v(1),...,v(n/2) are linearly independent, a(i) = 0 for each i between 1 and n/2.
now b(1)v(1)+b(2)v(2)+...+b(n/2)v(n/2) = 0 and for the same reason also b(i) = 0 for each i between 1 and n/2.
therefore B=(u(1),...,u(n/2),v(1),...,v(n/2)) is linearly independent and therefore a basis of V. and because for each i T(v(i)) = u(i), then [T(v(i))]B = ei
so B is your desired basis.
Sorry for my bad English.