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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 $\operatorname{Im}[T]=\ker(T)=0$?

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Hint: By the rank-nullity theorem,

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

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  • $\begingroup$ Thank you, all I needed to prove this question. $\endgroup$ – user95087 Oct 19 '13 at 1:49

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