Suppose $W$ is finite-dimensional and $T \in \mathcal L(V,W)$. Prove that $T$ is injective if and only if there exists $S \in \mathcal L(W,V)$ such that $ST$ is the identity map on $V$.

Suppose $T$ is injective and define $S(Tv) = v$. If $Tv_1=Tv_2$ then (by injectivity) $v_1=v_2$ meaning $S$ maps the same input to the same output thus it's a function (aka it's well defined).

It's easy to see $S$ is linear (it "inherits" linearity from $T$)

• $S(Tv_1 + Tv_2) = S(T(v_1+v_2)) = v_1+v_2$
• $S(\lambda Tv) = S(T(\lambda v)) = \lambda v$

Currently $S$ is only defined on the range of $T$, we can extend it to be defined on the rest of $W$ using 2.34 (this is where we need $W$ to be finite-dimensional!): If $W = \text{range }T \oplus U$ and $S$ is defined on $\text{range }T$ we're free to define $Su = 0$ for all $u \in U$. Now $S \in \mathcal L(W,V)$ as desired.

For the backwards direction suppose $ST$ is the identity map, then $Tv_1=Tv_2$ implies $S(Tv_1)=S(Tv_2) = v_1 = v_2$ so $T$ is injective.

More generally for any function the existence of a left inverse is equivilant to injectivity. The interesting thing here is that the linear structure was preserved.

Old bad proof

Note that $\text{null }T = \{0\}$ is finite-dimensional, and $\text{range }T \subseteq W$ is finite-dimensional, so 3b/16 implies $V$ is finite-dimensional and thus by 3.22

Let $w_1,\dots,w_r$ be a basis for $\text{range }T$, since each $w_j$ is in the range of $T$ we can let $v_1,\dots,v_r$ be a list in $V$ such that $w_j = Tv_j$, then define $Sw_j = v_j$ for all $1 \le j \le r$. Use 2.33 to extend to a basis $w_1,\dots,w_m$ of $W$ and finish definining $S$ by setting $Sw_j = 0$ for all $r < j \le m$.
The list $v_1,\dots,v_r$ is independent, because

Since $\dim \text{range }T = \dim V = r$ 2.39 implies the list $v_1,\dots,v_r$ forms a basis for $V$. Now can finally show $ST$ is the identity!

This completes the forward direction.

Now suppose $S$ exists such that $ST$ is the identity map on $V$. Clearly $\text{null }T \subseteq \text{null }ST = \{0\}$ so $T$ is injective by 3.16. This completes the backwards direction.

This took me ~1h, way too long! I diden't realize that $V$ coulden't be infinite-dimensional since it would force $\text{null }T$ to be infinite-dimensional, since $\text{range }T \subseteq W$ is clearly finite-dimensional.

I'm proud of this solution though, my first solution was to define $S$ as the inverse of $T$, then show it's well defined and linear. I like this solution much better since linearity is obvious (we just defined $S$ in terms of $W$'s basis).
edit: what the hell ^ this is garbage why was I ever proud of this, the only new thing here is linear structure being preserved. left inverses existing being the same as injectivity is true for all functions