# To show that injectivity of a linear map implies left-invertibility under the assumption that the target space is finite-dimensional

Let $T:V \rightarrow W$ be a linear map where the vector space $W$ is finite-dimensional.

$T$ is said to be injective iff for $v_1,v_2 \in V$ such that $T(v_1)=T(v_2)$, we always have $v_1=v_2$.

$T$ is said to be left-invertible iff there exists a linear map $S:W \rightarrow V$ such that $S\{T(v)\}=v$ for all $v \in V$.

We have to show that

$T$ is injective $\implies$ $T$ is left-invertible (under the assumption that $W$ is finite-dimensional)

I cannot penetrate the problem without the extra assumption of surjectivity, which I must not use here (of course, injectivity$~+~$surjectivity $\Longleftrightarrow$ invertibility). How do I utilize the assumption that $W$ is finite-dimensional ? Any help would be greatly appreciated.

• Do you know that there is a subspace $W_1$ of $W$ such that $W = W_1 \oplus T(V)$? If so you can define $S(x + T(y)) = y$ whenever $x \in W_1$ and $y \in V$. – user414998 Feb 11 '17 at 22:12

Pick a basis $A = \{v_1, \dots v_n\}$ for $V$. By injectivity of $T$, the set $B = \{T(v_1), \dots, T(v_n)\}$ is a basis for $T(V)$ in $W$. Extend $B$ to a basis $B'$ for all of $W$, so $B' = B \cup \{w_{n+1}, \dots w_{n+k}\}$.
Now define $S \colon W \to V$ by $S(T(v_i)) = v_i$, $S(w_j) = 0$, and extend by linearity.
There are details here for you to flesh out (like why is $V$ finite dimensional, why is $B$ a basis, why is $S$ well defined), but this is the general idea.