Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Digital systems, linear algebra, introduction to formal proof resit / 4 / b

Let $c_1=(1,1,1{)}^T$ and $c_2=(2,-2,6{)}^T$. The remaining columns are $3c_1$ and $2c_2$, so these two vectors span the column space. They are linearly independent: the determinant of their first two coordinates is $-4\ne 0$. Therefore $\{(1,1,1{)}^T,(2,-2,6{)}^T\}$ is a basis of the range, and the matrix rank is two.

Equivalently the range is the plane $y_3=2y_1-y_2$: both displayed basis vectors satisfy this equation, and their two-dimensional span fills that plane.

The independent-column representation gives $T(x)=c_1(x_1+3x_3)+c_2(x_2+2x_4)$. Thus a vector lies in the kernel of a linear map represented by this matrix precisely when $x_1=-3x_3$ and $x_2=-2x_4$. Writing the two free coordinates separately givesThe two displayed vectors are linearly independent, so they form a basis of the kernel. Its dimension two also agrees with the rank-nullity theorem, $4=2+2$.