# Homework Help: Non-degenerate Hermitian Matrices and their Eigenvalues

1. Nov 29, 2015

### RJLiberator

1. The problem statement, all variables and given/known data
Is there a non-degenerate 2x2 matrix that has only real eigenvalues but is not Hermitian? (Either find such a matrix, or prove that it doesn't exist)

2. Relevant equations

3. The attempt at a solution

Here's my problem. I'm getting Contradicting results.

So, I found this 2x2 matrix:
\begin{matrix}
1 & 2 \\
3 & 2
\end{matrix}

This matrix has eigenvalues of 4 and -1.

This matrix is not hermitian as the hermitian representation of this matrix transposes it.

Therefore this matrix is a 2x2 matrix that is non-degenerate and is NOT hermitian and has ONLY real eigenvalues.

This should be the example that the question was looking for.

However, in my notes, I have the equation for eigenvalues being
eigenvalue = [(a+b)+/- (squareroot ((a-b)^2-4cd))]/2

where
\begin{matrix}
a & c \\
d & b
\end{matrix}

And if we take this representation, we get negative eigenvalues.

What is going on here?

2. Nov 30, 2015

### Samy_A

Are you sure about that equation?
I find, by expanding $\det(A-\lambda I)=0$ and solving the quadratic equation:
$\lambda=\frac{(a+b)\pm\sqrt{(a+b)²-4(ab-cd)}}{2}$

3. Nov 30, 2015

### RJLiberator

Yes, I agree with what you are saying, and I noticed that as well, however, if you simplify it further:

(a+b)^2 = a^2+b^2+2ab
So then, 2ab-4ab becomes -2ab.

so now we have a^2+b^2-2ab-4cd
but a^2+b^2-2ab is just (a-b)^2 and so we get
(a-b)^2-4cd.

which is why I am getting contradictory results.
What is going wrong?

4. Nov 30, 2015

### Samy_A

It should be $+4cd$ after the simplification.

5. Nov 30, 2015

### RJLiberator

Ah....yes, that was the source of my frustration.

So, it seems my example now holds up and has nothing contradicting it.

Thank you kindly.