Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics I / 1 / c

Here $\dim V=2$. By the rank-nullity theorem, $r(T)+n(T)=2$, and a dimension is a nonnegative integer. All three possibilities occur:The images of a linear map in these examples have dimensions $0,1,2$, respectively, while the corresponding kernel of a linear map is ${\mathbb{R}}^2$, $\{(0,y):y\in \mathbb{R}\}$ and $\{0\}$. The identity above excludes all other pairs.

If $TS={\mathrm{id}}_W$, every $w\in W$ would satisfy $w=T(Sw)$, so $T$ would be a surjective linear map. But its image of a linear map has dimension at most $\dim V=2$, whereas $\dim W=3$. Therefore no such $S$ exists. This is the obstruction to a right inverse: $T$ cannot reach every target vector.

The necessary and sufficient condition isFor necessity, if $ST={\mathrm{id}}_V$ and $v\in \ker T$, then $v=S(Tv)=S(0)=0$.

For sufficiency, let $e_1,e_2$ be a basis of $V$ and suppose the kernel of a linear map is trivial. The vectors $T{e}_1,T{e}_2$ are linearly independent, since $aT{e}_1+bT{e}_2=0$ implies $a{e}_1+b{e}_2\in \ker T$. Extend them to a basis $T{e}_1,T{e}_2,w_3$ of $W$. Define a linear map $S$ byThen $ST$ agrees with the identity operator on a basis, hence on all of $V$. Thus a left inverse exists precisely when $T$ is injective; this is the linear left inverse criterion.