Simultaneously diagonalizing two $3\times 3$ commuting Hermitian matrices. 
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 $U$.
 A: 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 $\Lambda$.
