Characteristic polynomial for A

Prove that the characteristic polynomial for $A = \begin{bmatrix} B & 0 \\ 0 & C \end{bmatrix}$, where $B$ and $C$ are square matrices, is the product of the characteristic polynomials of $B$ and $C$.

My attempt is the following. Let $A \in M_{n\times n}(\mathbb{F})$ the polynomial $f(t)=\det(A-\lambda I_n)$ is the characteristic polynomial for $A$. I´m not sure if what should be done is to prove that the determinant of a matrix $A$ is equal to the product of its eigenvalues?

$$A=\begin{bmatrix} B & 0 \\ 0 & C \end{bmatrix},$$

$$P_A(\lambda)=(A-\lambda I),$$ $$P_A(\lambda)=\det(\begin{bmatrix}B-\lambda & 0 \\ 0 & C - \lambda \end{bmatrix}),$$ $$P_A(\lambda)= (B-\lambda)(C-\lambda)-0,$$ $$=BC-B\lambda -C\lambda +\lambda^2$$ $$P_A(\lambda)=\lambda^2 -\lambda(B+C) + BC.$$

Now, if I have $B \in M_{n\times n}(\mathbb{F})$ the polynomial $f(t)=\det(B-\lambda I_n)$ is the characteristic polynomial for $B$ and the same for $C$, $C \in M_{n\times n}(\mathbb{F})$ where its characteristic polynomial is $f(t)=\det(C-\lambda I_n)$.

The product of its characteristic polynomials is:

$$(B-\lambda I_n)(C-\lambda I_n)$$ $$(B-\lambda)(C-\lambda)$$ $$=BC-B\lambda -C\lambda +\lambda^2$$

But what do I do with the identity?

Please let me know if what I did is correct for the problem.

• How about just using the definition of characteristic polynomial and determinant of block matrices? That should give you the answer immediately. – Arthur Jan 5 '18 at 10:18

There's an error at the 3rd line of your computation: $\;\begin{bmatrix}B-\lambda & 0 \\ 0 & C - \lambda \end{bmatrix}$ is inconsistent: you cannot subtract a number to a matrix. : they'll have size , say $p$ and $q$ Also, if $A$ is a square matrix of size $n$, $B$ and $C$ certainly are not: they'll have size, say $p$ and $q$ such that $p+q=n$.
A block diagonal square matrix (with block square matrices): $$M =\begin{bmatrix}P & 0 \\ 0 & Q \end{bmatrix}$$ has determinant: $$\det M=\det P\cdot\det Q.$$