Hello,

The problem is;

Prove that if $\omega$ is a cube root of $1$ ($\omega\neq 1$), then the matrices $\begin{pmatrix}0&1&0\\0&0&1\\1&0&0 \end{pmatrix}$ and $\begin{pmatrix}1&0&0\\0&\omega&0\\0&0&\omega^2 \end{pmatrix}$ are similar.

Since I want to use $A \sim B \implies \det(\lambda I - A)=\det(\lambda I - B)$ I calculated the characteristic polynomial of each matrix. For the first it is $(\lambda -1)^3$ and for the second it is $(\lambda-1)(\lambda -w)(\lambda-w^2)$. In the second case you can expand to get $$-λ^3 + λ^2 - λ \omega^3 + \omega^3 + λ^2 \omega^2 - λ \omega^2 + λ^2 \omega - λ \omega.$$ Which can be grouped like $\lambda^3-\lambda(\omega^2+\omega+1)+\lambda(\omega^2+\omega +\omega^3)-\omega^3$. Since $\omega^2+\omega+1=0$ you can show that $\omega^3=1$ and the characteristic polynomial of $B$ becomes $\lambda^3-1$ but then this is clearly not equal to $(\lambda -1)^3$ so that the two matrices are similar must be false.

I tryed a couple of other things but nothing too exciting. I'm not sure at all what went wrong.

The characteristic polynomial of the first matrix is

$p_1(\lambda) = 1 - \lambda^3$

similarly

$p_2(\lambda) = -\lambda ^3+\lambda ^2 \left(\omega ^2+\omega +1\right)-\lambda \omega \left(\omega ^2+\omega +1\right)+\omega ^3$

so we have the following set of equations for similarity

$0 = \omega^2 + \omega + 1$

$\omega^3 = 1$

solving the first gets us

$\omega =\dfrac{-1\pm i\sqrt{3}}{2}$

these are the two cube roots of $1$ other than $1$

and so the second equation is satisfied as well.

Ah OK I did not succesfully calculate a determinant that's easy to fix though.

Thanks.

The second matrix is the Jordan Canonical Form of the first matrix

Originally Posted by Idea

The second matrix is the Jordan Canonical Form of the first matrix

Yes that I think is true. I am having trouble showing it though.

So I know the eigenvalues go on the diagonal. And I guess it's no problem to show that each of $1$, $\omega$, and $\omega^2$ solve $1-\lambda^3$. So then I think I'm done and have only to write $\dots \begin{pmatrix}1&0&0\\0&w&0\\0&0&w^2\end{pmatrix} \blacksquare$

Most proofs involving "Jordanizing" some transformation involve some business about the minimal polynomial but that seems irrelevant as there are 3 eigenvalues hence the matrix is diagonable. Or is there more?

Originally Posted by bkbowser

Yes that I think is true. I am having trouble showing it though.

So I know the eigenvalues go on the diagonal. And I guess it's no problem to show that each of $1$, $\omega$, and $\omega^2$ solve $1-\lambda^3$. So then I think I'm done and have only to write $\dots \begin{pmatrix}1&0&0\\0&w&0\\0&0&w^2\end{pmatrix} \blacksquare$

Most proofs involving "Jordanizing" some transformation involve some business about the minimal polynomial but that seems irrelevant as there are 3 eigenvalues hence the matrix is diagonable. Or is there more?

yes. 3 distinct eigenvalues is sufficient to say that the original real matrix is diagonalizable over the complex numbers (since the eigenvalues are not all real)

but it is not diagonalizable over the real numbers.

OK. We have as some standing asumption that the underlying field is always algebraically closed. So even though the question is pretty ambiguous I'd just say both transformations are on $\mathbb{C}$ and that would be the end of everything then.