# Invertible linear operator

Let $L\colon \mathbb{R}^n\to\mathbb{R}^n$ be an invertible linear transformation and let $V\subset \mathbb{R}^n$ be a subspace with $L(V)\subseteq V$. Show that $L(V)=V$.

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This is your third post in a very short time span. Please try to better format your posts (and notice how your previous posts had to be edited by others). Please also ask questions rather than stating problems as orders. Unfortunately, your posting pattern gives the impression of throwing your homework on the web without prior thought, and this is generally frowned upon. –  Jonas Meyer Feb 12 '11 at 19:14
Also downvoted for the same reason as Jonas. Also the lack of homework tag here is bothersome. –  fdart17 Feb 12 '11 at 20:23
One way of thinking about this is the following: Note that if $L$ is injective, then $L$ maps independent sets to independent sets. It follows that $L$ maps a basis of $V$ to an independent set, call it $A$. The assumption gives us that this set $A$ is contained in $V$. Now use that $V$ is finite dimensional to conclude that $A$ must be a basis of $V$. From this you should see easily how to conclude. –  Andres Caicedo Feb 12 '11 at 22:05

If $L(V)\subsetneq V$, then $L|_V\colon V\to V$ is not surjective and hence not injective (because $V$ is a finite dimensional vector space). Therefore $L$ is not injective, which contradicts the assumtion that $L$ is invertible.