# If all the eigenvalues of an operator are real, is the operator Hermitian?

How do I prove or disprove the following statement?

The eigenvalues of an operator are all real if and only if the operator is Hermitian.

I know the proof in one way, that is, I know how to prove that if the operator is Hermitian, then the eigenvalues must be real. It's the other direction that I'm not sure.

• In my practice I encountered an differential operator (a Hamiltonian), which was Hermitian in a space of functions with a non trivial scalar product, like $(\psi,\phi) = \int \psi(x)\phi (x)\rho (x) dx$. This Hamiltonian could be split into two parts: $\hat H = \hat H_0 + \hat V$, where $\hat H_0$ was hermitian in another space - with a different scalar product ($\rho =1$). Naturally the "perturbation" operator $\hat V$ was not Hermitian in this new basis. Yet, the perturbation theory worked. Nov 8, 2011 at 10:45
• @Vladimir: this is very fasionable nowadays--- you should publish your analysis (if you haven't already)--- this is PT quantum mechanics. Do you have a reference or a better description of V? Nov 8, 2011 at 18:02
• @RonMaimon: Yes, I have published it. First, it was a preprint in Russian and then, two articles in Russian journals. I submitted an English version of it on arXiv: arxiv.org/abs/0906.3504 Nov 8, 2011 at 18:31

We shall assume that the vector space $V$ (where the linear operator $A:V\to V$ acts) is a complex vector space (as opposed to a real vector space), and that $V$ is equipped with a sesquilinear form $\langle\cdot,\cdot \rangle:V\times V \to \mathbb{C}$. (We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.)

Since OP says he already knows how to prove that a Hermitian operator $A$ has real eigenvalues, he is essentially asking

If all the eigenvalues $\lambda_i$ are real, is the operator $A$ Hermitian?

The answer is No, only if $A$ is diagonalizable in an orthonormal basis.

In other words, the eigenspaces $\ker(A-\lambda_i {\bf 1})\subseteq V$ should be mutually orthogonal and together span the whole vector space $V$.

A version of the Spectral Theorem says that $A$ is orthonormally diagonalizable iff $A$ is a normal operator.

• Comment on the last sentence (it is correct but my poor mind was confused for a bit): The word orthonormally diagonalizable is important, because if it is diagonalizable by a non-orthogonal matrix $P$ so that $A = P D P^-1$, then it need not be normal. Mar 30, 2021 at 3:05

It's well known that it isn't true. Here is a counterexample: $\begin{pmatrix}1&1\\0&1\end{pmatrix}$. This is not hermitian, but it has two real eigenvalues +1,+1. This example is not diagonalizable, so it isn't so interesting.

A diagonalizable example is easy to construct too, if the eigenvectors are not orthogonal to each other. Consider the matrix $\begin{pmatrix}100&3\\-2&234\end{pmatrix}$. This matrix has two real eigenvalues close to 100 and 234, since the small perturbation of the eigenvalue equation doesn't change the discriminant. But the matrix is not symmetric, so it is not Hermitian. In this case, you can define a different metric on the vector space, a different definition of orthogonal, that makes the matrix Hermitian. This is easy-- the matrix is diagonal in it's Eigenbasis, with real eigenvalues, if you declare that this basis is orthonormal, then the matrix becomes Hermitian.

If you have a diagonalizable matrix with real eigenvalues $E_i$, and the eigenvectors $V_i$ are orthogonal and form a complete set,

Then the matrix is given by

$$E_i \bar{V}_i^j V_i^k$$

This reconstructs a Hermitian matrix from the list of orthogonal real eigenvalues. A proper statement is that a diagonalizable matrix with real eigenvalues and a basis of eigenvectors defines a metric on the complex vector space where it becomes Hermitian. The proof is to declare that all the eigenvectors have zero inner product, and some positive norm.

In the subject of PT symmetric quantum mechanics, this construction defines the metric on Hilbert space from the energy eigenstates.

• How does handle this PT theory the fact that non-hermitian observables are no longer the generators of infinitesimal transformations? how they do made up for that loss? Nov 17, 2011 at 22:23
• I don't know, I am still learning this theory. Perhaps you can ask this as a full-fledged question. Nov 18, 2011 at 5:52
• I don't see how $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ has two eigenvalues +1. Apr 25, 2018 at 22:05
• @my2cts the characteristic polynomial has two roots, as usual these are the two eigenvalues. Feb 3, 2021 at 20:55
• @ComptonScattering It has only one eigenvalue, 1, since it has aonly one eigenvector (1,0). Still, all its eigenvalues are real and it is not Hermitian. A valid counterexample. Feb 4, 2021 at 1:00