Quick question about Hermitian operators

[dEdt]

If H is a Hermitian operator, then its eigenvalues are real. Is the converse true?

 

[micromass]

No. Take as counterexample the matrix \left(\begin{array}{cc} 0 & 1\\ 0 & 0 \end{array}\right).
However, if you know that the operator is normal (that is: if AA^*=A^*A), then it is true that it is Hermitian iff the eigenvalues are real.

 

[NegativeDept]

No. Take as counterexample the matrix \left(\begin{array}{cc} 0 & 1\\ 0 & 0 \end{array}\right).
However, if you know that the operator is normal (that is: if AA^*=A^*A), then it is true that it is Hermitian iff the eigenvalues are real.

Good counterexample!

A few more useful rules: a matrix $M$ has "good" eigenvectors if and only if it is normal. "Good" eigenvectors means it is possible to choose eigenvectors such that they form an orthonormal basis for whatever vector space $M$ acts on.

Equivalently, a matrix is normal if and only if it can be diagonalized by a unitary transformation:
$M = U D U^{-1}$
for some unitary matrix $U$ and diagonal matrix $D$ whose diagonal elements are eigenvalues of $M$.

Hermitian matrices are always normal, and their eigenvalues are always real. But if $M$ is abnormal, it might have real eigenvalues and "bad" eigenvectors which are not orthogonal. Here's an example:

$\left[\begin{array}{cc} 1 & 1 \\ 0 & 2 \\ \end{array}\right]$

This thing has eigenvalues 1 and 2, but its eigenvectors

$\left[\begin{array}{c} 1 \\ 0 \\ \end{array}\right] \quad \left[\begin{array}{c} 1 \\ 1 \\ \end{array}\right]$

are not orthogonal to each other.

The example micromass gave is even more badly-behaved: it is defective, which means it doesn't have enough linearly-independent eigenvectors to span the vector space. So we can't even form a "bad" basis using its eigenvectors.