Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 1 / 2C (Vectors and Matrices)

The $j$th column isAll three columns therefore lie in the two-dimensional span of $v$ and $1$, so

For $x=(x_1,x_2,x_3{)}^T$,The vectors $v$ and $1$ are linearly independent, and $a\ne 0$. Thus $Ax=0$ exactly whenSolving gives

Every column lies in $\mathrm{span}\{v,1\}$. Conversely, differences of two columns are nonzero multiples of $1$, while any one column together with $1$ recovers $v$. HenceandThis is an instance of the rank of a matrix with affine columns.

The rank-nullity theorem states that for a linear map with finite-dimensional domain,Here the domain is ${\mathbb{R}}^3$, and the preceding parts givewhich verifies the theorem.

The $j$th column of $B$ isThus the image is contained in $\mathrm{span}\{v,1\}$. Since $n>1$, $v$ is not constant; since $m>1$ and $a\ne 0$, two columns have a nonzero constant difference. The two spanning vectors therefore both lie in the image, so $\mathrm{rank}B=2$. By rank-nullity theorem,