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

For the linear map $T:V\to W$, its image of a linear map and kernel of a linear map areBoth contain the zero vector, since $T(0)=0$. If $y_1=T(v_1)$ and $y_2=T(v_2)$ belong to the image of a linear map, then for any real numbers $a,b$,Thus $\mathrm{im}T$ is a vector subspace of $W$. If $u_1,u_2\in \ker T$, thenso $\ker T$ is a vector subspace of $V$.

The rank of a linear map and nullity of a linear map areThe rank-nullity theorem states thatTo prove it, choose a basis $u_1,\dots ,u_k$ of the kernel of a linear map and use basis extension to obtain a basisof $V$. The vectors $T(v_1),\dots ,T(v_{n-k})$ span the image of a linear map: applying $T$ to an arbitrary linear combination of the displayed basis removes all the $u_i$ terms. They are also linearly independent. Indeed,so ${\sum }_j{a}_j{v}_j={\sum }_i{b}_i{u}_i$ for some scalars $b_i$. The linear independence of the extended basis forces every $a_j$ and $b_i$ to vanish. Consequently $r(T)=n-k$ and $n(T)=k$, proving the identity. The same argument covers an empty basis when one of these dimensions is zero.

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.