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I am trying to find the system of eigenvalues and corresponding normalized eigenvectors for the following Hermitian matrix:

$$\mathbf{H}=\begin{pmatrix}10 & 3i \\ -3i & 2\end{pmatrix}$$

I began by solving the equation $\det(\mathbf{H}-\lambda\mathbf{I}_{2\times2})$, giving:

$$\det(\mathbf{H}-\lambda\mathbf{I}_{2\times2})=\lambda^2-12\lambda+11 \implies \lambda_{1}=1,\, \lambda_{2}=11$$

We then find our corresponding eigenvectors as follows:

$$\begin{pmatrix}10 & 3i \\ -3i & 2\end{pmatrix}\begin{pmatrix}x_{1} \\ y_{1}\end{pmatrix}=\begin{pmatrix}x_{1} \\ y_{1}\end{pmatrix} \implies x_1=k,\, y_1=3ik \\ \begin{pmatrix}10 & 3i \\ -3i & 2\end{pmatrix}\begin{pmatrix}x_{2} \\ y_{2}\end{pmatrix}=11\begin{pmatrix}x_2 \\ y_2\end{pmatrix} \implies x_2 = k,\, y_2=-\frac{i k}{3}$$

Normalizing this we get:

$$\vec{v}^{1}=\frac{1}{\sqrt{10}}\begin{pmatrix}1 \\ 3i\end{pmatrix} \implies \langle\vec{v}^{1},\vec{v}^{1}\rangle = 1\\ \vec{v}^{2}=\frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ -i\end{pmatrix} \implies \langle\vec{v}^{2},\vec{v}^{2}\rangle=1$$

However, this would imply that $\mathbf{U}^{\dagger}\mathbf{H}\mathbf{U}=\mathbf{\Lambda}$, with:

$$\mathbf{U}=\begin{pmatrix}\frac{1}{\sqrt{10}} & \frac{1}{\sqrt{2}} \\ \frac{3i}{\sqrt{10}} & -\frac{i}{\sqrt{2}}\end{pmatrix} \text{ and } \mathbf{\Lambda}=\begin{pmatrix}1 & 0 \\ 0 & 11\end{pmatrix}$$

However, when actually computing $\mathbf{U}^{\dagger}\mathbf{H}\mathbf{U}$ I get something very different to $\mathbf{\Lambda}$ (it's not even diagonal). So I have clearly done something very wrong, but I'm unsure where the error lies?

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  • $\begingroup$ $$\vec{v}^{2}=\frac{1}{\sqrt{2}}\color{red}{\begin{pmatrix}1 \\ -i\end{pmatrix}} \implies \langle\vec{v}^{2},\vec{v}^{2}\rangle=1$$ Where did the red come from? $\endgroup$ – Git Gud Dec 16 '13 at 19:46
  • $\begingroup$ Something that I always find helps in situations like this is checking that I have the right eigenvalues/eigenvectors using wolfram!alpha. $\endgroup$ – Avi Steiner Dec 16 '13 at 19:51
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Looks to me like $v_1=(3i,1)\implies \|v_1\|=\sqrt{10}$ so $\displaystyle{v_1\over \|v_1\|}={1\over \sqrt{10}}(3i,1)$.

And $v_2=(-i,3)\implies \|v_2\|=\sqrt{10}$ so $\displaystyle{v_2\over \|v_2\|}={1\over \sqrt{10}}(-i,3)$.

Then $$U=\begin{bmatrix} v_1 & v_2\end{bmatrix}={1\over \sqrt{10}}\begin{bmatrix} 3i & -i\\ 1 & 3\end{bmatrix}$$ and $$U^*HU=\begin{bmatrix} 11 & 0\\ 0 & 1\end{bmatrix}.$$

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  • $\begingroup$ This implies my $\vec{v}^{1}$ is wrong as well? $\endgroup$ – Thomas Russell Dec 16 '13 at 21:04
  • $\begingroup$ Looks that way. My quick scratch work on that part yields $x_1+{i\over 3}y_1=0$ so take $x_1=1\implies y_1=-i/3$. Mathematica yields $x_1=3i$, $y_1=1$ which is consistent with what I had since the latter is just the former times $3i$. $\endgroup$ – JohnD Dec 16 '13 at 21:21
  • $\begingroup$ Did this resolve your question? $\endgroup$ – JohnD Dec 17 '13 at 14:33

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