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. $\endgroup$ May 2, 2015 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? $\endgroup$ May 2, 2015 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. $\endgroup$ May 2, 2015 at 7:35

1 Answer 1


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. :) $\endgroup$ May 5, 2015 at 2:51

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