Past exam of the mathematics course of the University of Cambridge / 2014 / ib / Paper 2 / 1G / Solution

For a linear map $T:V\to W$ with $V$ finite-dimensional, the rank-nullity theorem statesTo prove it, choose a basis $v_1,\dots ,v_k$ of the kernel of a linear map and use basis extension to obtain $v_1,\dots ,v_n$ as a basis of $V$. The vectors $T{v}_{k+1},\dots ,T{v}_n$ span the image of a linear map: applying $T$ to a basis expansion kills the first $k$ terms. They are also linearly independent. Indeed, if ${\sum }_{j>k}{a}_{j}T{v}_j=0$, then ${\sum }_{j>k}{a}_j{v}_j\in \ker T$, so it is a combination of $v_1,\dots ,v_k$. Independence of the full basis forces every $a_j=0$. Consequently $\dim \mathrm{im}T=n-k$, proving the theorem.

Here the image of a linear map has dimension $r\in \{0,1,2,3\}$, since it lies in ${\mathbb{R}}^3$. ThusEvery possibility occurs: for $0\le r\le 3$, map $(x_1,\dots ,x_5)$ to $(x_1,\dots ,x_r,0,\dots ,0)\in {\mathbb{R}}^3$. This map has rank $r$ and kernel dimension $5-r$.