= Solution
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 <linear dependence>[linearly dependent]. All coefficients of $t^j$ for $j>k$ therefore vanish, so
$$
\deg\det(A+tB)\leq k.
$$
If $k=n$, then $B$ is invertible and the coefficient of $t^n$ is $\det B\ne0$. Hence the degree is exactly $n$.
For a zero-polynomial example with $k=1$, take
$$
A=\begin{pmatrix}0&0\\0&0\end{pmatrix},
\qquad
B=\begin{pmatrix}1&0\\0&0\end{pmatrix}.
$$
Then $\operatorname{rank}B=1$, while $\det(A+tB)=0$ for every $t$.
Back to article page
