Eigenvalues of Self-Adjoint Operators

**Guy** · September 5th 2010, 11:56 AM

Hey guys,

Let $\langle , \rangle$ be an arbitrary inner product on $\mathbb{R}^n$ . Suppose $A$ is self adjointing with respect to this inner product. Prove that $A$ has real eigenvalues.

My instinct would be to do something like this: given an eigenvalue $\lambda$ , let $v$ be a corresponding eigenvector. Then
$\displaystyle \lambda \langle v, v\rangle = \langle \lambda v, v\rangle = \langle A v, v\rangle = \langle v, A v\rangle = \langle v, \lambda v\rangle = \bar{\lambda} \langle v, v\rangle $

The thing about this is that the inner product is defined on $\mathbb{R}^n$ , not $\mathbb{C}^n$ , and I'm having a hard time changing my thought process away from what I did above. I've been thinking maybe it should be $\mathbb{C}^n$ since maybe the question would be trivial otherwise. Any thoughts?

**tonio** · September 5th 2010, 01:38 PM

Originally Posted by Guy

Hey guys,

Let $\langle , \rangle$ be an arbitrary inner product on $\mathbb{R}^n$ . Suppose $A$ is self adjointing with respect to this inner product. Prove that $A$ has real eigenvalues.

My instinct would be to do something like this: given an eigenvalue $\lambda$ , let $v$ be a corresponding eigenvector. Then
$\displaystyle \lambda \langle v, v\rangle = \langle \lambda v, v\rangle = \langle A v, v\rangle = \langle v, A v\rangle = \langle v, \lambda v\rangle = \bar{\lambda} \langle v, v\rangle $

The thing about this is that the inner product is defined on $\mathbb{R}^n$ , not $\mathbb{C}^n$ , and I'm having a hard time changing my thought process away from what I did above. I've been thinking maybe it should be $\mathbb{C}^n$ since maybe the question would be trivial otherwise. Any thoughts?

If the inner product is defined on $\mathbb{R}^n$ then any existing eigenvalue is real, otherwise they meant $\mathbb{C}^n$ and what you did is correct.

Tonio

**HallsofIvy** · September 5th 2010, 05:56 PM

Since the set of real numbers is a subset of the complex numbers, even though <u, v> is defined to be a real number, you can think of it as a function from pairs of vectors to a complex number.

For example, $R^2$ with inner product <(a, b), (c, d)>= ac+ bd has inner product always real but the matrix $\begin{bmatrix}0 & 1\\ -1 & 0\end{bmatrix}$ has characteristic equation $\lambda^2+ 1= 0$ which has i and -i as roots. Since we can't have Av= iv in $R^2$ , the corresponding linear transformation does NOT have any eigevalues. But we can, temporarily switch to "over $C^n$ so that any linear transformation has eigenvalues, show, as you did, that, for a self-adjoint linear transformation, they must be real and so assert that any self-adjoint linear transformation on a vector space over the real numbers has eigenvalues.

**Guy** · September 5th 2010, 06:58 PM

Originally Posted by HallsofIvy

Since the set of real numbers is a subset of the complex numbers, even though <u, v> is defined to be a real number, you can think of it as a function from pairs of vectors to a complex number.

For example, $R^2$ with inner product <(a, b), (c, d)>= ac+ bd has inner product always real but the matrix $\begin{bmatrix}0 & 1\\ -1 & 0\end{bmatrix}$ has characteristic equation $\lambda^2+ 1= 0$ which has i and -i as roots. Since we can't have Av= iv in $R^2$ , the corresponding linear transformation does NOT have any eigevalues. But we can, temporarily switch to "over $C^n$ so that any linear transformation has eigenvalues, show, as you did, that, for a self-adjoint linear transformation, they must be real and so assert that any self-adjoint linear transformation on a vector space over the real numbers has eigenvalues.

Yeah, I get that. I guess my concern for this particular problem is that it's part of the build up to the Spectral Theorem. We need (counting multiplicities and including 0) $n$ eigenvalues for our self-adjoint matrix to make things work I think, and so if you have an inner product over R^n you want to show that this is the case for self adjoint matricies. If you can extend the inner product to C^n, like you can with the dot product, then it's fine. I just worry that there is an inner product that you can't "switch over" to C^n in the same way and that I wouldn't be able to show the Spectral Theorem if we are working in R^n with an inner product that can't be extended to C^n.

EDIT: I think I've shown that you can extend any inner product over R^n to one over C^n by taking
ComplexInnerProduct(a + bi, c + di) = <a, c> + <b, d> + i[<b, c> - <a, d>]
so I guess it's irrelevant.

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