2

I was reading this question, and the second sentence of the accepted answer (with 34 upvotes) is

"As a linear transformation of Euclidean vector space E is Hermite iff there exists an orthonormal basis of E consisting of all the eigenvectors of B"

I think I have heard this sort of theorem multiple times, but I don't think it's true. Let $U \in \mathbb{C}^{n \times n}$ be a unitary matrix, and $D \in \mathbb{C}^{n \times n}$ a diagonal matrix with non-real entries. Consider the matrix $$B = U D U^{*}.$$ $\mathbb{C}^n$ has a basis consisting of eigenvectors of $B$, but $$B ^* = U^*D^*U \not = U^* D U = B$$ so $B$ is not hermitian.

Is my counterexample correct? What would be a correct reformulation of the "theorem" above? Should it be:

A basis $(b_1, ..., b_n)$ of $\mathbb{C}^n$ is a set of eigenvectors for a matrix $B$ with real eigenvalues iff $B$ is hermitian?

Thank you very much.

Helix
  • 307
  • Many (if not most) authors define a "Euclidean vector space" as a finite-dimensional inner product space over $\mathbb R$. Thus the eigenvalues are real when the linear transformation has an eigenbasis. – user1551 Apr 13 '21 at 14:45
  • 1
    Your last formulation does not make sense. A correct version would be something like "$B \in \Bbb C^{n \times n}$ is Hermitian if and only if $B$ has real eigenvalues and an orthonormal basis of associated eigenvectors". – Ben Grossmann Apr 13 '21 at 14:47
  • @user1551 Oh so we are viewing $V = \mathbb{C}^n$ as a real vector space, so in my example the entries of $D$ (being non-real) are not actually eigenvalues of $B$, so $B$ doesn't have any eigenvalues, so it doesn't have any eigenvectors, so it doesn't contradict the first stated theorem, right? – Helix Apr 13 '21 at 15:06
  • @BenGrossmann Thanks. Yes I see now, the forward direction makes sense, but not the backwards one. – Helix Apr 13 '21 at 15:07
  • @Helix Sorry, my bad. I thought you were talking about some passage in a book. Technically, the result in the block quote is correct, because the term "Euclidean vector space" does usually refer to a finite-dimensional real inner product space. However, in the context of the linked question, the ground field is clearly complex. Therefore, the sentence in the block quote, while true, does not apply to the answer you've mentioned. The correct result to use, like Ben wrote, is that $B$ is Hermitian iff it has a real spectrum and an orthonormal eigenbasis. – user1551 Apr 13 '21 at 17:42
  • @user1551 Thanks, I think I understand – Helix Apr 13 '21 at 20:29

0 Answers0