Proving that self-adjoint commuting linear operators have a simultaneous orthonormal eigenbasis

Question

I am trying to prove an extension to the result given as a Lemma below. I would appreciate feedback, and pointers to other ways of proving it.

Theorem. Let $V$ be a finite dimensional vector space. If the self-adjoint operators $A_1,\dots,A_r:V\to V$ commute, then there exists a simultaneous orthonormal eigenbasis for $V$ for $A_1,\dots,A_r$.

Lemma. Let $V$ be a finite dimensional vector space. If the self-adjoint operators $A_1,\dots,A_r:V\to V$ commute, then there exists a simultaneous eigenbasis for $V$, for $A_1,\dots,A_r$, though not necessarily orthonormal.

Proof of Theorem. We prove it by induction on $n$. Suppose first that $n=1$. Then there is a vector $e\in V$ with $||e|| = 1$, which is an ON-basis for $V$, and it is an eigenvector to all $A_1,\dots,A_r$. So the statement is true when $n=1$. Suppose that the statement is true also for $n=k$, and consider the case when $n=k+1$.

Let $e_1$ be a vector in the simultaneous eigenbasis that exists by the Lemma. Without loss of generality we may assume that $||e_1||=1$. Now let $U=(\text{span}\{e_1\})^{\bot}$. Denote by $\lambda_1^j$ the eigenvalue of $A_j$ to which the eigenvector $e_1$ corresponds. Since $A_j$ is self-adjoint, we have for an arbitrary vector $u\in U$ that $$ 0 = \langle \lambda_1^je_1,u\rangle = \langle A_je_1,u\rangle = \langle e_1,A_j u\rangle, $$ which shows that $U$ is invariant under the operators $A_1,\dots,A_r$. So the restrictions $A_1\rvert_U,\dots,A_r\rvert_U$ are self-adjoint (since $A_1,\dots,A_r$ are self-adjoint) linear operators on $U$.

By the induction hypothesis there exists a simultaneous ON-eigenbasis $\{e_2,\dots,e_{k+1}\}$ for $U$, since $\dim U = k$. But these vectors are also eigenvectors of $A_1,\dots,A_r$, and they are all orthogonal to $e_1$. Hence $e_1,\dots,e_{k+1}$ is a simultaneous ON eigenbasis for $V$. $\square$

Answer 1

Since every self-adjoint operator is normal and that any eigenbasis of a normal operator is an orthogonal list, the theorem is a direct result of the lemma. (I mean, it can be much neater, but your proof is OK.)

Below is some of my thoughts about other proof of the theorem:

Lemma 1: The theorem is true if we have:

Let $V$ be a finite dimensional vector space. If the self-adjoint operators $A_1, A_2, \cdots, A_r$ commutes, then they share a common eigenvector.

proof: Induction on dim$V$. Let $u$ be the common eigenvector and let $U$ be the orthogonal complement of span$u$. By since span$u$ is invariant under $A_i$, $U$ is also invariant under $A_i$. Consider $A_i$ restricted on $U$: it is obvious that they do commutes with each other and dim$U$ is less than dim$V$. So we completes the proof by the inductive hypothesis.

Note that since the other eigenvectors are in $\mathrm{span}u^\perp$, they are all orthogonal to $u$. This means, the eigenvectors we obtained is a orthogonal list.

Lemma 2: Let $V$ be a finite dimensional vector space. The commuting operators $A_1, A_2, \cdots, A_r$ share a common eigenvector if $A_1$ has a eigenvector, say $A_1u = \lambda u$.

proof: We first induct on r.

• If $A_1$ is a scalar multiply of the identity operator: by inductive hypothesis, $A_2,A_3,\cdots,A_r$(r-1 one) has a common eigenvector, say $v$, clearly, $v$ is also an eigenvector of $A_1$.

• Otherwise $A_1$ is not a scalar multiply of the identity operator, which means ker$(A_1 - \lambda I)$ is not $V$, and its dim is less than dim$V$. Note that ker$(A_1 - \lambda I)$ is invariant under $A_1, A_2, \cdots, A_r$. So coniser $A_1, A_2, \cdots, A_r$ restricted on ker$(A_1 - \lambda I)$, by induct on dimension of $V$, we completes this bullet.

OK, I will omit the proof of the next lemma:

Lemma 3: Any self-adjoint operator has an eigenvector.

Putting together the three lemmas in a reverse order, one can easily see that the theorem is proved.

Remark:

• Since I only used the results from normal operator (except in lemma 3), the theorem shall also apply to normal operators if $V$ is a complex (On real vector space, the existence of eigenvector is not ensured).
• In fact, if all the eigenvalues are different, the existence of eigenvectors is ensured and all of the them are orthogonal, so it applies to commuting pairs among which one has no multiply eigenvalues.
• In fact, the theorem can be strengthen, see Simultaneously Diagonalizable Proof .

The proof of lemma 2 is adapted from Harm Derksen: The Fundamental Therorem of Algebra and Linear Algebra.

I hope there won't be too many mistakes in my arguments, but if any, comment below.

Last line: what do you mean by simultaneous? If this means if any one of the operator has a set of eigenbasis then all of the share the same set of eigenbasis, then, it apply to normal operator on both real and complex vector space.