1
$\begingroup$

This question already has an answer here:

Consider the following matrix:

\begin{bmatrix} -3 & 1 & -1 & & & & & \\ -7 & 5 & -1 & & 0 & & & 0\\ -6 & 6 & -2 & & & & & \\ & & & 4 & 0 & 1 & & \\ & 0 & & 0 & 1 & 0 & & 0\\ & & & 0 & 0 & 4 & & \\ & & & & & & -1 & -1\\ & 0 & & & 0 & & 1 & -3 \end{bmatrix}

I need to find its minimal polynomial.
I found that the characteristic polynomial is: $P(\lambda)=(\lambda+2)^4(\lambda-4)^3(1-\lambda)$ using the fact that for a block matrix:
$$A=\begin{bmatrix} {A}_{1} & 0 & \cdots & 0 \\ 0 & {A}_{2} & \cdots & 0 \\\vdots & \vdots & \ddots & \vdots \\0 & 0 & \cdots & {A}_{n} \end{bmatrix} $$ It holds that: $\det(A)=\prod_{i=1}^{n}\det(A_i)$

Is there an easy trick to find the minimal polynomial as well?

$\endgroup$

marked as duplicate by Alex M., hardmath, Martin Sleziak, Claude Leibovici, Jonas Meyer Jun 21 '15 at 5:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ It is the least common multiple of the minimal polynomials of the blocks. $\endgroup$ – Alex M. Jun 18 '15 at 9:47
3
$\begingroup$

$p(A)=0$ is equivalent to $p(A_1)=\dots=p(A_n)=0$

So it is the minimum common multiple of the blocks minimum polynomials.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.