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Assume $W$ is finite dimensional and $T:V\rightarrow W$ is linear.

Show that if $T$ is injective, then there exists a linear map $S:W\rightarrow V$ so that $ST=I_V$.

Can I have a hint on how to define $S$? If w = T(v), then I will define S(w) = v. But I'm not sure how to define S(w') for w' that are not hit by $T$. I feel like we need cases: (1) $w$ that are hit and (2) $w$ that are not hit.

EDIT: I've seen solutions in which people find the basis of W. What's the point in finding a basis? Can't we just say that $S(w)=v$ if $Tv=w$? That doesn't require us finding a basis, right? Or do we find a basis for the case in which there doesn't exist a $v$ such that $T(v)=w$?

I've noted my confusion in the comments below. Any help would be much appreciated.

beginner
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    That is already a good idea. One way of realizing this is to take a basis for the image of $T$ and extend it to one on $W$ and define $S$ appropiately on these basis vectors – leoli1 Feb 09 '21 at 16:52
  • Injective linear map has left inverse: See here or here or here. I hope you can do it now. – Sumanta Feb 09 '21 at 17:13
  • @leoli1 Thank you for the hint. On a related note, I'm trying to improve my understanding of linear transformation in general by proving additivity/homogeneity properties. So I'm trying to prove additivity in the case that $T$ is bijective. I define $S$ as $S(w) = v$. But I'm getting circular arguments: $S(w_1+w_2)=S(T(v_1)+T(v_2))=S(T(v_1+v_2)) = S(w_1+w_2)$. How do I break $S(w_1+w_2)$ into $S(w_1)+S(w_2)$? – beginner Feb 09 '21 at 17:20
  • @Mathlover Thank you. I will look at those once I become more familiar with simpler concepts such as my above comment. – beginner Feb 09 '21 at 17:22
  • You need to use the direct sum decomposition w.r.t. $T(V)$. See here – Sumanta Feb 09 '21 at 17:25
  • Prove the following steps: $(1)$ $T(V)$ is a vector subspace of $W$. $(2)$ The map $T':V\to T(V)$ defined by $T'(x)=T(x)$ for all $x\in V$ is a linear isomorphism. $(3)$ Choose a basis $\mathcal B'$ for $T(V)$ and then extends this basis to a basis $\mathcal B$ for $W$. $(4)$ For any $y\in W$ write $y=y'+z$ where $y'\in \text{span}(\mathcal B')$. $(5)$ Define $S:W\to V$ as $S(y)=T'^{-1}(y')$. Can you complete it now? – Sumanta Feb 09 '21 at 17:33
  • @Mathlover Regarding the T is injective problem, can you write up a solution? I've been trying to understand this for hours but haven't made any progress. I'm especially confused by the first two links you sent because both of those answers don't prove homogeneity/additivity. – beginner Feb 09 '21 at 19:59
  • @leoli1 What's the point in finding a basis? Can't we just say that $S(w)=v$ if $Tv=w$? That doesn't require us finding a basis, right? Or do we find a basis for the case in which there doesn't exist a $v$ such that $T(v)=w$? For this https://math.stackexchange.com/questions/2140011/to-show-that-injectivity-of-a-linear-map-implies-left-invertibility-under-the-as I'm wildly confused where $S(w_j)$ represents. Like is $w_j$ a basis vector? If it is, why is the user defining this function on a basis vector? – beginner Feb 09 '21 at 20:08
  • If we just want to define $S$ on the image of $T$, this will be no problem and we can do that without a basis. However in order to show that we can extend this $S$ to the whole of $W$ we need a basis or some related argument (since the statement is wrong if we replace vector spaces with for example modules over arbitrary rings) – leoli1 Feb 09 '21 at 20:14
  • @leoli1 Okay, I'm on board with that. What's confusing to me though is how finding a basis for the whole of $W$ useful for trying to define $S$ for $w \in W$ where there doesn't exist a $v$ such that $T(v) = w$. – beginner Feb 09 '21 at 20:19

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The image of $T$ is a subspace $U=T(V)$ of $W$.

On this subspace, define $S(w)=v$ if $T(v)=w$.

Since $T$ is injective, $S$ is a mapping from $U$ into $V$

Then one can extend this mapping to all of $W$ by setting $S(w)=0$ if $w\not\in U$.

Wuestenfux
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    A point not in $U$ might have a part in $T(V)$ i.e. you need to consider $W=U\oplus U'$ and then define $S(w)=T^{-1}(w_1)$ where $w=w_1+w_2$ for $w_1\in U$ and $w_2\in U'$. Please edit your answer. – Sumanta Feb 09 '21 at 17:06