Question

Is there anyone who could explain to me how to find the eigenvalues of a Hermitian matrix?

Answer 1

For example,\[A=\left[\begin{matrix}-3 & 2+i & 0 \\ 2-i & -3&0\\0&0&3\end{matrix}\right]\]

Answer 2

You calculate them the same you find any eigenvalues. Knowing that A is Hermitian tells you eigenvalues must necessarily be real; so you can check your calculations that way.

Answer 3

For your particular A, you ahve two diagonal blocks and we can write down the characteristic polynomial as the product of the chr poly for those blocks, namely:

\[[(\lambda+3)^2 - (2^2 + 1^2)](\lambda-3) = (\lambda^2 + 6\lambda+4)(\lambda-3)\]

Answer 4

Really sorry, but could you expand that a bit by telling me how you came to that answer? I'm particularly confused about how to manipulate/incorporate/use/what to do with the 'i' terms... (Or more the case, the 2+-i terms)... Cheers!

Answer 5

I really appreciate the help! I've been combing the internet and all of my lecture notes on how to do it but I can't seem to get it right :(

Answer 6

let me know when you're back and we can chat

Answer 7

Right.... I think I got it... Can you tell me if this is right? :\[(2-\lambda)\det\left(\begin{matrix}-3-\lambda&2+i \\ 2-i & -3-\lambda\end{matrix}\right)\]\[=(3-\lambda)[(-3-\lambda)(-3-\lambda)-(2+i)(2-i)]\]\[=(3-\lambda)(9+3\lambda+3\lambda+\lambda^2-4-i^2)\]
Since i = sqrt(-1), does that mean that the i^2 term becomes -1? ie\[=(3-\lambda)(4+6\lambda+\lambda^2)\]Also, if the -2i and 2i terms had not cancelled each other, for example if I was left with an 'i' term, how does this affect the eigenvalues?
Cheers for the help :) I appreciate it!

Answer 8

Yes, you've got the calculation.

The point is for a Hermitian matrix is that the 'i' terms will always cancel out, because the essential property of such matrices is that their eigenvalues are real; so we we certainly can't have any i appearing in the characteristic equation.

However, there's nothing wrong with complex eigenvalues per se.