1
$\begingroup$

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 $U$.

$\endgroup$
  • 2
    $\begingroup$ 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
0
$\begingroup$

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 $\Lambda$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.