I have been given a problem a while ago with no given solution so I thought I'd check with the folks at mathstackexchange to see if my method is correct. It goes like this:
I have been given a random matrix $ A \in \mathbb{R}^{4\times4}$ with the following property:
- The characteristic polynomial is $\chi_A(t) = m_A(t)(t-1)(t+1)$
I have to prove that every matrix with this property has the same Jordan Normal Form and to also find it.
My attempt : It's quite obvious from the characteristic polynomial that the eigenvalues are $1$ and $-1$. Since the minimal polynomial shares the same roots as the characteristic polynomial (a.k.a the eigenvalues) I have concluded that the minimal polynomial has to be $m_A(t) = (t-1)(t+1)$ for every matrix of this type and so it follows that $\chi(t) = (t-1)^2(t+1)^2$
Then from the minimal polynomial I can tell based on the power of each factor ($(t-1)$ and $(t+1)$) that the biggest Jordan block for each eigenvalue has the size $1$
and therefore I get $J_A = \begin{pmatrix}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& -1& 0\\0& 0& 0& -1 \end{pmatrix}$
Is this correct? Did I skip any step or use redundant information anywhere?
While it is clear that, having factors $t-1$ and $t+1$, the characteristic, polynomial must have roots $1$ and $-1$, it is not obvious that it cannot have any other roots. It can be proved, using the fact that $\deg(\chi_A)=4$, and that all roots of the characteristic polynomial are also roots of the minimal polynomial, so $t^2-1$ divides $m_A$ and therefore $\deg(m_A)\geq2$; the degrees now leave no room for any other factors. However, the conclusion would not hold if the matrix could be square of a larger size (so some argument using the size is necessary). With $m_A$ having simple roots, $A$ is diagonalisable and the diagonal form is also a Jordan form, and the multiplicities of the diagonal entries are those as root of $\chi_A$, so your conclusion is correct.