Showing two matrices have the same minimal polynomial

Question

Show that the matrices $\begin{pmatrix}-1&0&0&0\\0&-1&0&0\\0&0&2&0\\0&0&0&-1\end{pmatrix}$ and $\begin{pmatrix}2&0&0&0\\0&2&0&0\\0&0&2&0\\0&0&0&-1\end{pmatrix}$ have the same minimal polynomials.

Is it correct to say the characteristic polynomials of the two are ($x-2$)($x+1$$)^3$ and ($x-2$$)^3$($x+1$) respectivly and the minimal polynomials are ($x-2)(x+1$) since that's the irreducible factorization of both?

Answer 1

The minimal polynomial of $A$ is the smallest monic polynomial $p$ (by degree) such that $p(A) = 0$. So you suspect that $(x-2)(x+1) = x^2 -x -2$ is the minimal polynomial of each of those matrices (say name them $A$ and $B$). One way to check that would be to

1. Calculate $A^2$ and $B^2$, and check that $ A^2 -A -2 = 0$ and $ B^2 -B -2 = 0$
2. Make a brief argument that a linear polynomial cannot be the minimal polynomial of either $A$ or $B$.

Answer 2

Let $A=\begin{pmatrix}-1&0&0&0\\0&-1&0&0\\0&0&2&0\\0&0&0&-1\end{pmatrix}$ and $B=\begin{pmatrix}2&0&0&0\\0&2&0&0\\0&0&2&0\\0&0&0&-1\end{pmatrix}$.

we know that the dimension of the eigen space: $dim(E_\lambda)=n-rank(A-\lambda I)$, for a $n\times n$ matrix $A$. So $dim(E_{-1})=4-rank(A+ I)=3$. So the number of Jordan block corresponding to the eigenvalue $=3$. Since algebraic multiplicity of $-1$ is $3$, so highest order Jordan block of $-1$ is $1$. Similarly for the eigenvalue $2$ highest order Jordan block is $1$. So the minimal polynomial of $A$ is $m_A(t)=(t+1)(t-2)$.

Also for the matrix $B$, $dim(E_{-2})=4-rank(A+ 2I)=3$ and by same reason $m_B(t)=(t+1)(t-2)$. Although their characteristic polynomial are different.