# 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 $$U$$.

• 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 $E_k$. – MoonLightSyzygy Jan 2 '20 at 0:11
• 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. – Math1000 Jan 2 '20 at 0:29
• The sign in my $\Omega E_k=E_k$ is a $\subset$, which is what follows from the proof in the next sentence. – MoonLightSyzygy Jan 2 '20 at 0:31
• Oh, I suppose $(1,0,1)$ and $(-1,2,1)$ are eigenvectors of $\Omega$ with eigenvalue zero then? – Math1000 Jan 2 '20 at 0:32
• If you got that their images are $0$, they should be. – 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 $$\Lambda$$.