Undergrad A strange definition for Hermitian operator

[struggling_student]

In lecture notes at a university (I'd rather not say which university) the following definition for Hermitian is given:

An operator is Hermitian if and only if it has real eigenvalues.

I find it questionable because I thought that non-Hermitian operators can sometimes have real eigenvalues. We can correctly say that Hermitian operators can only have real eigenvalues but that does not define the operator, right? Is it some kind of convention or is it just plain wrong? Alas the physicists often don't understand the difference between an implication and equivalence.

 

[dextercioby]

The statement which was give to you is wrong. One can find a non-hermitean matrix with real eigenvalues.

 

[Twigg]

Counterexample: $$
\begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix} $$
has eigenvalue 1 with multiplicty 2. It's not Hermitian.

 

[hilbert2]

A matrix is hermitian if it has real eigenvalues and you can diagonalize it with a unitary transformation. This means that if and only if matrix $A$ is hermitian, there exists a matrix $U$ such that $U^\dagger U = UU^\dagger = 1$ and $U^\dagger A U$ is a diagonal matrix with real numbers on the diagonal.