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

Let $V$ be an $n$-dimensional real vector space and $T:V\rightarrow V$ a linear transformation from $V$ to itself. Suppose that $T[V]=\ker(T)$. Show that $n$ is even.

I'm really lost. Since $\operatorname{Im}[T]=\ker(T)$, does it mean that there has to be an even number of $T(\vec v)$ such that half of them are additive inverse of the other half, and $\operatorname{Im}[T]=\ker(T)=0$?

$$n = \dim V = \dim T[V] + \dim \ker(T)$$