Given $$H = \begin{pmatrix}2&3-i\\3+i&-1\end{pmatrix}$$ find a unitary matrix $U$ such that $$U^\dagger H U = D$$ where $D$ is the diagonal matrix containing the eigenvalues of $H$.

I assumed $U$ would therefore be the eigenvector matrix of $H$ as that normally gives a diagonal matrix containing the eigenvalues. I found the eigenvalues to be $\lambda = 4,-3$ and the respective eigenvectors to be

$$\begin{pmatrix}\frac{3-i}{2}\\1\end{pmatrix}, \quad \begin{pmatrix}\frac{i-3}{5}\\1\end{pmatrix}$$

However the eigenvector matrix isn't unitary. How do I find another matrix that fits the relationship, but is unitary?


After normalising the eigenvector matrix the diagonal matrix doesnt give the original eigenvalues, but a scaled down version:

enter image description here

  • $\begingroup$ You might start by normalizing the two eigenvectors that you found. Remember: there’s no such thing as the eigenvectors. Any nonzero scalar multiple of an eigenvector is also one. $\endgroup$ – amd Apr 30 at 6:27

You have done everything correct except for the small mistake in the last step. While normalising the vector, you have multiplied by the norm instead of dividing it and hence you are getting different values. For the first column, instead of $\frac{\sqrt{14}}{2}$ the term should be $\frac{2}{\sqrt{14}}$. And similarly for the other column. That should fix the error.


You have correctly calculated two eigenvectors of $H$.

In order to produce an orthonormal basis of eigenvectors (to get the corresponding unitary matrix), you want your eigenvectors $v^{(1)}, v^{(2)}$ to satisfy:

$$\left \langle v^{(1)}, v^{(2)} \right \rangle := v^{(1)}_1 \overline{v^{(2)}_1} + v^{(1)}_2 \overline{v^{(2)}_2} = 0 \ \ \ \text{ and} $$ $$ \left\|v^{(1)}\right\| = \left\|v^{(2)}\right\| = 1$$

Your current eigenvectors already satisfy the first equation (they are orthogonal). All you need to do is to scale them by the appropriate constants so that their norm becomes $1$.

That is, can you find $c_1, c_2$ such that $\left\|c_1\begin{pmatrix} \frac{3-i}{2}\\ 1 \end{pmatrix}\right\| = \left\|c_2\begin{pmatrix} \frac{i-3}{5}\\ 1 \end{pmatrix}\right\| = 1$?

  • $\begingroup$ Could you please explain how the first equation represents the two vectors being orthogonal, or send a link that does. $\endgroup$ – Vishal Jain Apr 30 at 6:40
  • $\begingroup$ The definition of the standard inner product on $\mathbb{C}^n$ is $\left \langle v,w \right \rangle = \sum \limits_{i=1}^{n}v_i\overline{w_i}$. (In our case, of course n=2). Orthogonality is defined as $\left \langle v,w \right \rangle = 0.$ These are the only two facts that I am using in the first equation. $\endgroup$ – tia Apr 30 at 7:14
  • 2
    $\begingroup$ Being comfortable with inner products and their basic properties is an essential prerequisite for working with unitary matrices. I would suggest reviewing this topic before moving further in your linear algebra studies. $\endgroup$ – tia Apr 30 at 7:18
  • $\begingroup$ After normalising each eigen vector and doing the computation I got a diagonal matrix with eigen values that were both smaller by a factor of 4.9. Is there anyway way to see why changing the size of the eigen vector affects the eigen values in this way? I would've thought they were independent of the eigen vectors magnitude. Have I just done it wrong? My c1 and c2 were $\frac{\sqrt{14}}{2}$ and $\frac{\sqrt{35}}{5}$ respectively. $\endgroup$ – Vishal Jain Apr 30 at 9:26
  • $\begingroup$ The normalization constants are probably correct, but the eigenvalues are not supposed to change. Would you mind sharing your computation in more detail? $\endgroup$ – tia Apr 30 at 16:04

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.