# Existence of a linear transformation

Suppose that $V$ and $W$ are finite dimensional vector spaces and that $T:V \rightarrow W$ is injective. Prove than there is a linear transformation $S:W \rightarrow V$ such that $(S \circ T)(\overrightarrow v)=\overrightarrow v$ for all $\overrightarrow v$ in $V$.

At first sight, I thought I would have to show that T is bijective and therefore invertible which would give us $S=T^{-1}$. But there is not enough given to show that.

Our professor proved the linear extension theorem last time and used it to solve this problem:

Suppose $T$ is injective Let {$\overrightarrow v_{1},...,\overrightarrow v_{n}$} be a basis for $V$

Since T is injective, $T(\overrightarrow v_{1}),...,T(\overrightarrow v_{n})$ are linearly independent vectors in W

Let $\overrightarrow w_{1}=T(\overrightarrow v_{1})$ ... $\overrightarrow w_{n}=T(\overrightarrow v_{n})$ and extend the set to a basis for W {$\overrightarrow w_{1},...,\overrightarrow w_{m}$}

Define $S:W\rightarrow V$ as

$S(\overrightarrow w_{1})=\overrightarrow v_{1}$ ... $S(\overrightarrow w_{n})=\overrightarrow v_{n}$, $S(\overrightarrow w_{n+1})=\overrightarrow 0 ... S(\overrightarrow w_{m})=\overrightarrow 0$

By construction: $S$ is linear and $S(T(\overrightarrow v_{1}))=S(\overrightarrow w_{1})=\overrightarrow v_{1}$ ... $S(T(\overrightarrow v_{n}))=S(\overrightarrow v_{n})=\overrightarrow v_{n}$

$\therefore \exists$ $S:W\rightarrow V$ such that $S\circ T=id_{V}$

I have the following questions: How does T being injective serve us in this problem? How is it possible that S exists since T is only injective and not bijective? Why are the values of of the transformation by S equal to $\overrightarrow 0$ after $\overrightarrow w_{n+1}$? What does it mean to say "by construction, S is linear"?

If $T$ were not injective you couldn't prove that the image of a basis was linearly independent.

A left inverse $S$ exists for any injective function $T$. It will be a right inverse also if and only if $T$ is surjective as well.

The values of $S$ are $0$ for the extra basis vectors because you chose to define $S$ that way. Any other values would do as well.

You defined $S$ by saying what it did on a basis, and then used the fact that the basis spans to define what it does everywhere by assuming linearity. That is, your construction of $S$ makes it linear.

• A worst-case scenario: If $T(x) = 0$ for all $x$, would it still be true that $T(v_1)$, $T(v_2)$, \ldots, $T(v_n)$ are linearly independent? – Catalin Zara Mar 25 '16 at 19:58
• Thanks for the answer. Where are we using the fact that T is injective in the proof? The professor just states it and says that the images are linearly independent but I don't see it used in defining S. – Omrane Mar 25 '16 at 19:58
• @CatalinZara, no because in that case $KerT=V \not =$ {$\overrightarrow 0$} – Omrane Mar 25 '16 at 20:00
• My friend @CatalinZara has pointed you to the answer to your question. – Ethan Bolker Mar 25 '16 at 20:00
• Exactly. You can't extend a non linearly independent set of vectors to a basis. – Catalin Zara Mar 25 '16 at 20:02