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

Gaussian elimination givesThe first two original columns form an image basis:Their independence follows already from their third and first coordinates. The third column is twice the first plus the second, and the fourth is the first plus twice the second, so they span every output of $T$.

The rank-nullity theorem states $\dim V=\dim \ker T+\dim \mathrm{im}T$ for a linear map from a finite-dimensional vector space $V$. Here the domain has dimension four and the image has dimension two. Thus $\dim \ker T=4-2=2$. For example, a kernel basis is $\{(-2,-1,1,0),(-1,-2,0,1)\}$, as follows from the reduced equations.

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 $T$ 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.

A linear map is one-to-one, or injective, when equal outputs imply equal inputs. Equivalently its kernel is $\{0\}$: subtracting inputs converts equality of outputs to a zero output. Here $T$ is not one-to-one, since its kernel has dimension two. Concretely, $T(-2,-1,1,0)=T(0)=0$ for distinct inputs.