linear algebra - rank-nullity

**worc3247** · December 24th 2010, 01:01 PM

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)

**Drexel28** · December 24th 2010, 02:30 PM

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)

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