A <linear map> $T:V\to W$ is onto, or <surjective>, when every $w\in W$ equals $T(v)$ for some $v\in V$; equivalently its <image> is all of $W$. Here \b[$T$ is not onto], because its image has dimension two while its declared codomain $\mathbb R^3$ has dimension three. Indeed every image vector satisfies $x+y-3z=0$, so, for example, $(0,0,1)$ is not attained.

Solution

Past exam of the year 1 of the computer science course of the University of Oxford / 2018 / Digital systems, linear algebra, introduction to formal proof / 5 / c / iii / Solution

 Solution (source code)