Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 3 / 9E (Linear Algebra)

The rank-nullity theorem says that for a linear map $T:V\to Z$ with finite-dimensional domain,

The direct sum $U\oplus W$ consists of pairs $(u,w)$ with componentwise vector-space operations. The canonical inclusions areFor subspaces $U,W\le V$, defineThis is surjective, andThe First isomorphism theorem for vector spaces therefore gives

Row reduction of the matrices whose columns are the displayed spanning vectors givesRow-reducing the matrix formed from all six columns givesThe dimension formula for a sum of subspaces now yields

Expand $\det (A+tB)$ by multilinearity of the determinant in its columns. The coefficient of $t^j$ is a sum of determinants obtained by choosing $j$ columns from $B$ and the remaining columns from $A$. Since the columns of $B$ span a space of dimension $k$, every choice of more than $k$ columns from $B$ is linearly dependent. All coefficients of $t^j$ for $j>k$ therefore vanish, so

If $k=n$, then $B$ is invertible and the coefficient of $t^n$ is $\det B\ne 0$. Hence the degree is exactly $n$.

For a zero-polynomial example with $k=1$, takeThen $\mathrm{rank}B=1$, while $\det (A+tB)=0$ for every $t$.