# If $T:V\to W$ is an injective linear map, show there is a linear map $L:W\to V$ such that $LT=1_V$ [duplicate]

Let $V$ and $W$ be finite dimensional vector spaces over a field $F$. Let $T:V\to W$ be a linear transformation. Suppose that $T$ is one-to-one. Show that there is a linear transformation $L:W\to V$ such that $LT=1_V$.

So far I have: Since $T$ is injective, then a basis $\{v_1,\ldots,v_n\}$ for $V$ gets sent to a linearly independent set $\{T(v_1),\ldots,T(v_n)\}$ of $W$. I'm not sure how to expand the basis for $W$.

Is it just $B_W=\{w_1,\ldots,w_m\}$?

I know that if $T$ is 1-1, then $\dim V\leq\dim W$. So $n\leq\dim W$.

Any help from here would be great.

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## marked as duplicate by Jim, Aang, Andreas Caranti, Davide Giraudo, Dennis GulkoMar 9 '13 at 9:12

Hint 1: If you have a linearly independent set of vectors, you can add more vectors to that set to get a basis. Hint 2: If you want to define a linear transformation (L, in this case), it is enough to say where it sends each basis vector. –  Alex Zorn Mar 9 '13 at 7:52
You can find some good starting points on how to format mathematics on the site here. This AMS reference is very useful. –  Zev Chonoles Mar 9 '13 at 7:54
If you need clarification please edit or comment on your question instead of starting a new one. –  Jim Mar 9 '13 at 8:03
we can define L:W->V. For any w in the image of T, so there is a v in V with T(v)=w. We can define L(w)=v. The transformation T is one-to-one so v is unique. –  kkkk Mar 9 '13 at 8:09

Maybe a different solution than most would expect. As $T$ is injective we know that $T^T T$ is a quadratic Matrix with full rank, so it is invertible. Now define $$L=(T^T T)^{-1} T^T.$$ Than $LT=(T^T T)^{-1} T^T T=I_V$

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