Is it true that in a finite-dimensional inner product space over $\mathbb R$, every self-adjoint operator has an eigenvalue?

I know this is true in a complex vector space, but can't seem to see why it should hold in a real vector space, given that in general, a linear operator need not have any eigenvalues at all in the latter case.

  • $\begingroup$ The key is self-adjointness here. That implies that all zeros of the characteristic polynomial are real. Once you've proved that, you're done. $\endgroup$ – Daniel Fischer May 2 '15 at 7:16
  • $\begingroup$ I did prove that if a self-adjoint operator does have an eigenvalue, it has to be real. But the characteristic polynomial needn't split in $\mathbb R$, implying that eigenvalues needn't exist in the first place. Does self-adjointness somehow make the characteristic polynomial split? $\endgroup$ – Train Heartnet May 2 '15 at 7:24
  • $\begingroup$ It does, but that comes after. First see that a self-adjoint operator has at least one eigenvalue. Now, you don't know yet that the characteristic polynomial splits over $\mathbb{R}$, but what you do know is that it has at least one zero in $\mathbb{C}$. Now use self-adjointness to see that all complex zeros of the c.p. are real, hence (real) eigenvalues. $\endgroup$ – Daniel Fischer May 2 '15 at 7:35

Yes, every self-adjoint operator $T$ on a (non-trivial) finite-dimensional real inner product space $V$ has an eigenvalue.

The quickest way to obtain the result is to look at the complexification $V_{\mathbb{C}}$ and the operator $T_{\mathbb{C}}$ induced by $T$ there. Then $T_{\mathbb{C}}$ is self-adjoint, hence all its eigenvalues are real. If $\lambda$ is an eigenvalue of $T_{\mathbb{C}}$, and $z = u + iv$ is an eigenvector of $T_{\mathbb{C}}$ for the eigenvalue $\lambda$, i.e.

$$T_{\mathbb{C}}(u+iv) = Tu + iTv = \lambda(u+iv),$$

then the real and imaginary parts of $z$ are eigenvectors of $T$ or $0$. Since $z\neq 0$, not both of $u$ and $v$ can be $0$, so $\lambda$ is an eigenvalue of $T$.

We can also obtain the result by purely real methods:

The unit sphere $S_V = \{ v\in V : \lVert v\rVert = 1\}$ is compact, and the map $q\colon v \mapsto \langle v, Tv\rangle$ is continuous. Hence $q$ attains its maximum on $S_V$, say at $v_0$. For every $w \in S_V \cap (\operatorname{span} \{v_0\})^\perp$ and $\alpha \in \bigl[-\frac{\pi}{2},\frac{\pi}{2}\bigr]$, we have

\begin{align} f(\alpha) &:= \bigl\langle (\cos \alpha )v_0 + (\sin \alpha)w, T\bigl((\cos\alpha)v_0 + (\sin\alpha)w\bigr)\bigr\rangle\\ &= \cos^2\alpha \langle v_0, Tv_0\rangle + 2 \sin\alpha\cos\alpha\langle w,Tv_0\rangle + \sin^2\alpha \langle w,Tw\rangle, \end{align}

and by assumption $f$ has a local maximum at $\alpha = 0$, so

$$0 = f'(0) = 2\langle w, Tv_0\rangle,$$

whence $Tv_0 \in (\operatorname{span} \{v_0\})^{\perp\perp}$, i.e. $v_0$ is an eigenvector of $T$.

Once the existence of an eigenvalue is established, it follows that self-adjoint operators are orthogonally diagonalisable, i.e. there are orthonormal bases of $V$ consisting of eigenvectors of the self-adjoint operator $T$:

If $W\subset V$ is a $T$-invariant subspace, then $W^\perp$ is also $T$-invariant: Let $w\in W$ and $x\in W^\perp$. Then

$$\langle w, Tx\rangle = \langle Tw, x\rangle = 0,\tag{1}$$

where the first equality is due to the self-adjointness of $T$ and the second due to the $T$-invariance of $W$, whence $Tw\in W$ for $w\in W$. Since $w$ was arbitrary in $(1)$, it follows that $Tx\in W^\perp$. Since $x$ was arbitrary, we have $T(W^\perp)\subset W^\perp$.

Once we have one eigenvector $v_0$ of $T$, we consider the subspace $W = (\operatorname{span} \{v_0\})^\perp$. By the above, $W$ is $T$-invariant, so we can examine the operator $T_W \colon W \to W$ obtained by restriction. $T_W$ is again self-adjoint, hence unless $W = \{0\}$ has an eigenvalue. And $\dim W < \dim V$, so the construction ends after finitely many steps.

  • $\begingroup$ Oh, I understand now! Thank you so much for the help. :) $\endgroup$ – Train Heartnet May 5 '15 at 2:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.