characteristic polynomial of block matrix

Question

How do I prove that the characteristic polynomial of a block matrix p(x) of A: $$\begin{pmatrix}A_1&...0&...0\\ ...0&A_2&...0\\ ...0&...0&A_n\end{pmatrix}$$ is the multiplication of the characteristic polynomials of each of the block matrices: $$\Pi ^n_{i=1}p_i\left(x\right)$$

and the minimal polynomial of A is the lcm (least common multiple) of the block matrices ?

Maybe I can use somehow minimal polynomials and invariant sub-spaces according to fact that before that I proved that:

$$m_{T|W}\:\left(x\right)|m_T\:\left(x\right),$$

where m is the minimal polynomial.

Answer 1

Suppose $A_i$ is a $\nu_i \times \nu_i$ matrix.We apply the generaized Laplace expansion of $\det(\lambda I-A)$ along the top $\nu_1$ rows, then we do another generalized Laplace expansion along the top $\nu_2$ rows and so on, finally giving $$\det(\lambda I-A)=\det(\lambda I_{\nu_1}-A_1)...\det(\lambda I_{\nu_n}-A_n)$$ Let the minimal polynomial of $A_i$ be $f_i.$ Let $f=\text{ lcm }(f_1, ... ,f_n)$. Let $g$ be the minimal of $A.$ Then g(A)=0, so $g(A_1)=0,...,g(A_n)=0$ so $f_1|g, ...,f_n|g$ and thus $f|g.$Conversely $f(A_1)=0, ..., f(A_n)=0$,so $f(A)=0$ and thus $g|f$. Hence $f=g.$