# Is it a bijection between spaces?

Let $f:V\to V$, where $V$ is a finite dimensional vector space. Is it true that if $\operatorname{im}f=V$ then $f$ is a bijection?

-
I assume $f$ is supposed to be a linear map? –  Ben Millwood Dec 14 '12 at 3:06

Yes. It is consequence of the Dimension theorem.

-

Yes, and in fact more can be said: $\,f\,$ is a bijection iff it is surjective iff it is injective.

-

Choose a basis of $V$, say $v_1 \dots v_n$. Then since $f$ is surjective, $f(v_1) \dots f(v_n)$ must span $V$. Hence they must be linearly independent (since any linear dependence would allow us to throw out one of the vectors and get the same span, but $n-1$ vectors can't span $V$). Hence $f(v_1)\dots f(v_n)$ is in fact a basis for $V$.

Suppose $v$ is some vector with $f(v)=0$. Write $v$ in components wrt the basis $v_i$, say $v=\sum \lambda_i v_i$. Then $f(v) = f(\sum \lambda_i v_i) = \sum \lambda_i f(v_i)$ by linearity. But $f(v)=0$, so since the $f(v_i)$ are linearly independent, $\lambda_i = 0$ for all $i$. But that means $v = 0$, so $f$ is injective.

-