I am currently working through Sheldon Axler's Linear Algebra Done Right (3rd Edition), but am stumped on how to complete an exercise. The one in question is #21 from section 3B which states:
"Suppose $V$ is finite-dimensional and T $\in \mathcal{L}(V,W)$. Prove that $T$ is surjective if and only if there exists $S \in \mathcal{L}(W,V)$ such that $TS$ is the identity map on $W$".
My attempt so far:
Suppose $T$ is surjective. By the Fundamental Theorem of Linear Maps, $\dim V=\dim\operatorname{null}T+ \dim\operatorname{range}T$. Since $T$ is surjective, we have that $\operatorname{range}T$=$W$. Combined with the fact that $V$ is finite dimensional, we also have that $W$ (and thus range $T$) is also finite dimensional. Thus, for any $w \in W$, we have $w=Tv$ for $v \in V$. Define $Sw=v$ so that $TSw=Tv=w$, meaning that $TS$ is the identity map on $W$.
Now, let $TS$ be the identity map on $W$. For arbitrary $w \in W$, we have $Sw=v$ for some $v \in V$ and that $Tv=w$. Hence, for all $w \in W$ there is $v \in V$ such that $w=Tv$, which means range $T$=$W$ making $T$ surjective.
$\square$
My main concern was about how I defined $S$ in the forward direction. If $T$ is not injective, then I thought it might be possible to have $Sw$ map to more than one output. Could anyone provide an explanation on whether my line of thinking is correct or not? Additionally, if there are any other errors in the proof, please let me know. Thank you!
Best Answer
Unfortunately, your $S$ is not well-defined because $Sw$ maps to more than one output (as yourself have pointed out). Moreover, you have not proved that $S\in\mathcal L(W,V)$.
$(\implies)$ Since a linear transformation to a larger dimensional space can not be surjective, then $W$ must be finite-dimensional. Suppose $\dim W=n$. Let $\{w_i\vert i=1,2,\ldots,n\}$ be a basis of $W$. Since $T$ is sujective, for each $w_i$ there exists $v_{ij_i}\in V$ such that $$(*) \ \ \ \ \ \ w_i=T(v_{ij_i}),$$ where $j_i=1,2,\ldots,m$. Here $m$ denotes the number of vectors in $V$ which are mapped to $w_i$. For example, if there are three vectors in $V$ which are mapped to $w_2$, then $j_2=1,2,3$ and, therefore, $w_2=T(v_{21}),w_2=T(v_{22})$, and $w_2=T(v_{23})$.
Now, define $S:W\to V$ such that $$(**) \ \ \ \ \ \ S(w_i)=v_{i1},$$ where $i=1,2,\ldots,n$. Thus, $S\in\mathcal L(W,V)$.
Let $w\in W$ be arbitrary, then it follows from $(*)$ and $(**)$ that \begin{align} TS(w)&=TS(a_1w_1+a_2w_2+\cdots+a_nw_n)\\ &=T(a_1S(w_1)+a_2S(w_2)+\cdots+a_nS(w_n))\\ &=T(a_1v_{11}+a_2v_{21}+\cdots+a_nv_{n1})\\ &=a_1T(v_{11})+a_2T(v_{21})+\cdots+a_nT(v_{n1})\\ &=a_1w_1+a_2w_2+\cdots+a_nw_n\\ &=w. \end{align} Therefore, $TS=\operatorname{Id}_W.$