Exercise 5.1.21

If the first row of $B$ is all zero, then $\det <!--\; FUNCTION\; APPLICATION\; -->({\widetilde{A-\mathit{tB}}}_{1j})$ is a polynomial with degree less than or equal to $k$ and ${(A-\mathit{tB})}_{1j}$ contains no $t$ for all $j$. If the first row of $B$ is not all zero, then $\det <!--\; FUNCTION\; APPLICATION\; -->({\widetilde{A-\mathit{tB}}}_{1j})$ is a polynomial with degree less than or equal to $k-1$ and each ${(A-\mathit{tB})}_{1j}$ contains $t$ with degree at most $1$. In both case, we get that $\det <!--\; FUNCTION\; APPLICATION\; -->(A-\mathit{tB})$ is a polynomial with degree less than or equal to $k$.

Now we may induct on $n$ to prove the original statement. For $n=1$, we have $f(t)=A_{11}-t$. For $n=2$, we have $f(t)=(A_{11}-t)(A_{22}-t)-A_{12}{A}_{21}$. Suppose the hypothesis is true for $n=k-1$. For the case $n=k$, we expand the determinant along the first row. That is,

By the induction hypothesis, we know that

where $p(t)$ is a polynomial with degree less than or equal to $k-3$, and

is a polynomial with degree less than or equal to $k-2$. So it becomes

in which the summation of the second term and the third term is a polynomial with degree less than or equal to $n-1$.

and it would be