# Prove that if $T[V]=\ker(T)$, then $n$ is even, where $\dim(V)=n$

Prove that if $T[V]=\ker(T)$, then $n$ is even, where $\dim(V)=n$

We know that from the dimension thereom:

rank(T) + nullity(T) = n

So we can say that $n=\dim(T[V])+\dim(\ker(T))$

But I don;t know how to go on from here. What exactly does it mean when it says: $T[V]=\ker(T)$

I am aware there is a similar question to this already on MSE, but the answer didn't provide details and so I'm finding it hard to understand:

Let $V$ be $n$ dimensional real vector space. Show that of $T[V]=\ker(T)$, then $n$ is even

The image is the same as the kernel so in particular the rank and nullity of $T$ are equal.
• but why does this imply $n$ even Oct 25, 2017 at 23:54
• @KSplitX: $x + x = 2x$
Combine your $n=\dim(T[V])+\dim(\ker(T))$ and $T[V]=\ker(T)$ to get $$n=\dim(T[V])+\dim(T[V])=2\dim(T(V))$$
so $n$ is even.