# Thread: linear algebra - rank-nullity

1. ## linear algebra - rank-nullity

Let V be an n-dimensional vector space where $n\geq 1$. If $T: V\rightarrow V$is a linear transformation, prove that the following statements are equivalent:
a) im(T)=ker(T)
b) $T^2 = 0$, n is even and $r(T)=\frac{n}{2}$.

So far I have managed show show that (a) implies (b), but when trying to find (b) implies (a) I get stuck. This is my working so far:

dim(V) = r(T) + n(T)
$r(T) = \frac{n}{2}, dim(V) = n,$ so $r(T)=n(T)=\frac{n}{2}$

$T^2(V)=0, T(T(V))=0$, so T(V) $\epsilon$ ker(T)

2. Originally Posted by worc3247
Let V be an n-dimensional vector space where $n\geq 1$. If $T: V\rightarrow V$is a linear transformation, prove that the following statements are equivalent:
a) im(T)=ker(T)
b) $T^2 = 0$, n is even and $r(T)=\frac{n}{2}$.

So far I have managed show show that (a) implies (b), but when trying to find (b) implies (a) I get stuck. This is my working so far:

dim(V) = r(T) + n(T)
$r(T) = \frac{n}{2}, dim(V) = n,$ so $r(T)=n(T)=\frac{n}{2}$

$T^2(V)=0, T(T(V))=0$, so T(V) $\epsilon$ ker(T)
You're almost there. YOu know that since $\dim \ker T=\frac{n}{2}$ and $n$ is even that $\dim T(V)=\frac{n}{2}$. But, as you noticed since $T^2=\bold{0}$ we have that $T(V)\subseteq \ker T$. But, since $\dim T(V)=\dim \ker T$ this implies that $T(V)=\ker T$ (a subspace of a finite dimensional space which has the same dimension of the ambient space must be the full space)