a block matrix proof about characteristic polynomials

Question

If

$\hspace{2in}$$ A = \begin{bmatrix} B & 0 \\ C & D \end{bmatrix} \in \mathsf{M}_n, $

where $ B \in \mathsf{M}_k $ and $ D \in \mathsf{M}_{n-k}$, prove that $ p_A = p_B p_D $. ${Hint}$: proceed by induction on $n$ and expand the determinant across the first row.

i have no idea what to do. All i know is that $p_A (t) = \det(tI_n-A)$ , $p_B (t) =\det(tI_n-B)$ and that $p_D(t) = \det(tI_{n-k}-D) $

i also feel like you can prove this without induction by saying that $\det(A) = BC$

but i also feel like that is totally incorrect

What should i do? how do i prove this?

if you have a better title feel free to chage it

how would induction even play into this?

Answer 1

The easiest trick to implement (but not necessarily to think of) is to write $tI - A$ as the product $$ tI - A = \pmatrix{tI - B & 0\\0&I}\pmatrix{I & 0\\-C&I} \pmatrix{I & 0\\0&tI - D} $$ and it suffices to determine that the matrices in this product have determinants $\det(tI - B),1,\det(tI - D)$ (in that order). We could prove those formulas using induction, if you like. In particular, the formulas for $$ \det \pmatrix{I & 0\\0&tI - D}, \det\pmatrix{I & 0\\C&I} $$ are very nicely proven using the hint as it's given.

Answer 2

This is really not about characteristic polynomials at all, just a fundamental property of determinants (over any commutative ring $R$, where here we take $R$ to be the ring of polynomials in $t$ over your field), namely $$ \det\pmatrix{B&0\\C&D}=\det(B)\det(D). $$ You can apply this immediately for the characteristic polynomial, since the act of transforming $A$ into $xI_n-A$ amounts to transforming $B$ into $tI_k-A$, and $D$ into $xI_{n-k}-D$ (also $C$ becomes $-C$).

That property of determinants is the subject of this other question, and in my opinion the best proof is really directly from the (Leibniz formula) definition of determinants, as I detailed in my answer to that question. In particular, I would want to avoid using a property like $\det(XY)=\det(X)\det(Y)$, which although of course true, is actually quite a bit harder to prove directly from the definition.