About this determinant $|\lambda I − B(I + vu)|$

Question

$B$ is a complex $n\times n$ matrix. $u$ is $1 \times n$ and $v$ is $n \times 1$.

I have the following determinant: $|\lambda I − B(I + vu)|$.

According to my textbook, if $\lambda$ divides this determinant, then $B$ or $I + vu$ is singular. Why is that the case?

Thank you!

Answer 1

First of all, the determinant $$f(\lambda)=|\lambda I-B(I+vu)|$$ is a polynomial in $\lambda$. If $\lambda$ divides $f(\lambda)$, then $\lambda$ is a factor of $f(\lambda)$, i.e. $0$ is a root for $f(\lambda)$. Hence $$f(0)=|B(I+vu)|=|B||I+vu|=0$$ At least one of $|B|$ or $|I+vu|$ is $0$. Now recall that a square matrix is nonsingular iff it has a nonzero determinant and you have your conclusion.

Answer 2

Your determinant is the characteristic polynomial of $B(I+vu)$. The assumption means that its constant term is $0$. But hte constant term of the characteristic polynomial is the determinant , up to a sign. Hence $\det(B(I+vu))=0=\det(B)\det(I+vu)$.

Thus, $\det(B)=0$ or $\det(I+vu)=0$ and we are done.