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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.

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  • $\begingroup$ How about just using the definition of characteristic polynomial and determinant of block matrices? That should give you the answer immediately. $\endgroup$ – Arthur Jan 5 '18 at 10:18
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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 $p+q=n$.

Hint:

A block diagonal square matrix (with block square matrices): $$M =\begin{bmatrix}P & 0 \\ 0 & Q \end{bmatrix}$$ has determinant: $$\det M=\det P\cdot\det Q.$$

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