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


Hint: By the rank-nullity theorem,

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.