Let $$ \Omega = \begin{bmatrix}1&0&1\\0&0&0\\1&0&1\end{bmatrix},\quad \Lambda = \begin{bmatrix}2&1&1\\1&0&-1\\1&-1&2\end{bmatrix}. $$

By considering the commutator, show that these matrices may be simultaneously diagonalized. Find the eigenvectors common to both and verify that under a unitary transformation to this basis, both matrices are diagonalized.

Clearly the commutator $[\Omega,\Lambda]=0$ because the matrices commute (as can be checked by computing $\Omega\Lambda$ and $\Lambda\Omega$). Now, I computed the characteristic polynomial of $\Omega$ as $$ p_\Omega(\lambda) = \lambda^2(2-\lambda) $$ which has roots $\lambda=0$, $\lambda=0$, and $\lambda=2$, and the characteristic polynomial of $\Lambda$ as $$ p_\Lambda(\lambda) = (2-\lambda)(\lambda-3)(\lambda+1) $$ which has roots $\lambda=2$, $\lambda = 3$, and $\lambda = -1$. So $\Omega$ is degenerate and $\Lambda$ is not. But I don't see how these matrices have common eigenvectors, and am unsure as to what unitary transformation to the basis of common eigenvectors would simultaneously diagonalize both matrices. Any advice?

Edit: Okay, so according to @MoonLightSyzygy's hint, we have that the eigenvectors of $\Lambda$: $(1,1,-1)$, $(1,0,1)$ and $(-1,2,1)$ are also eigenvectors of $\Omega$. But what would be the unitary transformation to this basis under which both matrices are diagonalized? I know it is of the form $U$ where $U^*U=I$ and $U^*$ denotes the adjoint of $U$.

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    $\begingroup$ Compute the eigenspaces $E_2,E_3, E_{-1}$ of $\Lambda$, check that $\Omega E_k=E_k$ and therefore, eigenvectors of $\Lambda$ are also eigenvectors of $\Omega$. This is a general property since if $\Lambda v=rv$, then $\Lambda\Omega v=\Omega\Lambda v=r\Omega v$. Take a basis formed by non-zero vectors from each $E_k$. $\endgroup$ – MoonLightSyzygy Jan 2 '20 at 0:11
  • $\begingroup$ Hmm, I get $E_2 = \{c((1,1,-1):c\in\mathbb R\}$, $E_3 = \{c(1,0,1):c\in\mathbb R\}$, and $E_{-1}= \{c(-1,2,1):c\in\mathbb R\}$. But $\Omega E_2 = E_2$ while $\Omega E_3 = \Omega E_{-1} = \{(0,0,0)\}$. So something is wrong. $\endgroup$ – Math1000 Jan 2 '20 at 0:29
  • $\begingroup$ The sign in my $\Omega E_k=E_k$ is a $\subset$, which is what follows from the proof in the next sentence. $\endgroup$ – MoonLightSyzygy Jan 2 '20 at 0:31
  • $\begingroup$ Oh, I suppose $(1,0,1)$ and $(-1,2,1)$ are eigenvectors of $\Omega$ with eigenvalue zero then? $\endgroup$ – Math1000 Jan 2 '20 at 0:32
  • $\begingroup$ If you got that their images are $0$, they should be. $\endgroup$ – MoonLightSyzygy Jan 2 '20 at 0:33

Solving the eigenvector equation for $\Lambda$ and $\lambda=-1$ yields $(1,-2,-1)$; normalizing, we have $\frac1{\sqrt 6}(1,-2,-1)$. For $\lambda=2$ we have $(1,1,-1)$; normalizing we have $\frac1{\sqrt 3}(1,1,-1)$. For $\lambda = 3$ we have $(1,0,1)$; normalizing, we have $\frac1{\sqrt 2}(1,0,1)$. Since $\Lambda $ is non-degenerate and $[\Omega,\Lambda]=0$ we must have that these are eigenvalues for $\Omega$ as well; indeed \begin{align} \begin{bmatrix}1&0&1\\0&0&0\\1&0&1\end{bmatrix}\begin{bmatrix}1\\-2\\-1\end{bmatrix} &= \begin{bmatrix}0\\0\\0\end{bmatrix}\\ \begin{bmatrix}1&0&1\\0&0&0\\1&0&1\end{bmatrix}\begin{bmatrix}1\\1\\-1\end{bmatrix} &= \begin{bmatrix}0\\0\\0\end{bmatrix}\\ \begin{bmatrix}1&0&1\\0&0&0\\1&0&1\end{bmatrix}\begin{bmatrix}1\\0\\1 \end{bmatrix} &= \begin{bmatrix}2\\0\\2\end{bmatrix}, \end{align} corresponding to the eigenvalues $0$, $0$, and $2$ of $\Omega$. The columns of the unitary transformation are given by the normalized eigenvectors: $$ U=\begin{bmatrix} \frac1{\sqrt3}&\frac1{\sqrt6}&\frac1{\sqrt 2}\\ \frac1{\sqrt3}&-\frac2{\sqrt6}&0\\ -\frac1{\sqrt3}&-\frac1{\sqrt6}&\frac1{\sqrt 2}\\ \end{bmatrix} $$ and indeed we may compute $$ U^*\Lambda U = \begin{bmatrix}\frac1{\sqrt3}&\frac1{\sqrt3}&-\frac1{\sqrt3}\\\frac1{\sqrt 6}&-\frac2{\sqrt 6}&-\frac1{\sqrt6}\\\frac1{\sqrt 2}&0&\frac1{\sqrt2}\end{bmatrix} \begin{bmatrix}2&1&1\\1&0&-1\\1&-1&2\end{bmatrix}\begin{bmatrix} \frac1{\sqrt3}&\frac1{\sqrt6}&\frac1{\sqrt 2}\\ \frac1{\sqrt3}&-\frac2{\sqrt6}&0\\ -\frac1{\sqrt3}&-\frac1{\sqrt6}&\frac1{\sqrt 2}\\ \end{bmatrix} = \begin{bmatrix}2&0&0\\0&-1&0\\0&0&3\end{bmatrix} $$ and $$ U^*\Omega U = \begin{bmatrix}\frac1{\sqrt3}&\frac1{\sqrt3}&-\frac1{\sqrt3}\\\frac1{\sqrt 6}&-\frac2{\sqrt 6}&-\frac1{\sqrt6}\\\frac1{\sqrt 2}&0&\frac1{\sqrt2}\end{bmatrix} \begin{bmatrix}1&0&1\\0&0&0\\1&0&1\end{bmatrix}\begin{bmatrix} \frac1{\sqrt3}&\frac1{\sqrt6}&\frac1{\sqrt 2}\\ \frac1{\sqrt3}&-\frac2{\sqrt6}&0\\ -\frac1{\sqrt3}&-\frac1{\sqrt6}&\frac1{\sqrt 2}\\ \end{bmatrix} = \begin{bmatrix}0&0&0\\0&0&0\\0&0&2\end{bmatrix}. $$ It follows that $U$ simultaneously diagonalizes $\Omega$ and $\Lambda$.


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