📂Linear Algebra

Simultaneous Eigenvector

Definition¹

Let $V$ be a vector space, and let $\mathcal{A}$ be a collection of linear operators on $V$.

$$ \mathcal{A} = \left\{ A : V \to V \mid A \text{ is linear}\right\} \subset \operatorname{End}(V) $$

A nonzero vector $v \in V$ satisfying the following is called a simultaneous eigenvector for $\mathcal{A}$.

$$ A v = \lambda_{A} v, \quad \forall A \in \mathcal{A} $$

The $\lambda_{A}$’s are called simultaneous eigenvalues.

Explanation

A simultaneous eigenvector is, just as the name says, a vector that is an eigenvector for several linear operators at the same time. The corresponding eigenvalue may differ depending on the operator $A$. By the theorem below, operators that commute with one another on a finite-dimensional complex vector space have a simultaneous eigenvector. Operators that do not commute with one another generally do not have a simultaneous eigenvector.

The theorem is stated for three operators, but the same conclusion holds more generally for a collection $\mathcal{A}$ of mutually commuting linear operators. The condition of being a finite-dimensional complex vector space in the theorem is likewise needed to guarantee that each operator has an eigenvalue. Moreover, the converse holds. That is, two operators having a simultaneous eigenvector commute. This is easy to show: let $v$ be a simultaneous eigenvector of $A$ and $B$.

$$ \begin{align*} (AB - BA)v &= (A \lambda_{B} - B\lambda_{A})v \\ &= (\lambda_{B}A - \lambda_{A}B)v \\ &= (\lambda_{B}\lambda_{A} - \lambda_{A}\lambda_{B})v \\ &= 0 \end{align*} $$

The extension of simultaneous eigenvalues to linear functionals is called a weight.

Theorem

Suppose we are given a finite-dimensional complex vector space $V$ other than $\left\{ 0 \right\}$ and linear operators $A$, $B$, $C$ on $V$. If $A$, $B$, $C$ commute with one another, then there exists at least one simultaneous eigenvector for $\mathcal{A} = \left\{ A, B, C \right\}$.

Proof

Suppose $A$, $B$, $C$ commute with one another.

$$ [A, B] = AB - BA = 0 = [B, C] = [C, A] $$

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Part 1. The eigenspace of $A$

Since $V$ is a finite-dimensional complex vector space other than $\left\{ 0 \right\}$, the characteristic polynomial of $A$ is a polynomial with complex coefficients of degree $\dim V \ge 1$, and by the fundamental theorem of algebra it has a root. Let one of those roots be $\lambda_{A} \in \mathbb{C}$. Since the roots of the characteristic polynomial are eigenvalues, $\lambda_{A}$ is an eigenvalue of $A$, and let the eigenspace of $A$ corresponding to it be given as follows.

$$ V_{A} := \left\{ v \in V : Av = \lambda_{A}v \right\} $$

Since $\lambda_{A}$ is an eigenvalue, $V_{A} \ne \left\{ 0 \right\}$, and $V_{A}$ is a subspace of $V$.

Part 2. $V_{A}$ is $B$-invariant and $C$-invariant

Let $v \in V_{A}$. Since $A$ and $B$ commute, the following holds.

$$ \begin{align*} A(Bv) &= (AB)v \\ &= (BA)v \\ &= B(Av) \\ &= B(\lambda_{A}v) \\ &= \lambda_{A}(Bv) \end{align*} $$

Since $A(Bv) = \lambda_{A}(Bv)$, we have $Bv \in V_{A}$, and therefore $V_{A}$ is a $B$-invariant subspace. Since $A$ and $C$ also commute, the same computation shows that $V_{A}$ is $C$-invariant as well.

Part 3. Narrowing down to the eigenspace of $B$

Since $V_{A}$ is $B$-invariant, we may consider the restriction $B|_{V_{A}} : V_{A} \to V_{A}$. Since $V_{A}$ is also a finite-dimensional complex vector space other than $\left\{ 0 \right\}$, for the same reason as in Part 1, $B|_{V_{A}}$ has an eigenvalue $\lambda_{B} \in \mathbb{C}$. If we set the eigenspace of $B|_{V_{A}}$ corresponding to it as follows, then $V_{AB} \ne \left\{ 0 \right\}$.

$$ V_{AB} := \left\{ v \in V_{A} : Bv = \lambda_{B}v \right\} $$

$V_{AB}$ is $C$-invariant. Indeed, if $v \in V_{AB}$, then by Part 2 we have $Cv \in V_{A}$, and since $B$ and $C$ commute, the following holds.

$$ B(Cv) = (BC)v = (CB)v = C(Bv) = C(\lambda_{B}v) = \lambda_{B}(Cv) $$

Therefore $Cv \in V_{AB}$.

Part 4. Simultaneous eigenvector

Since $C|_{V_{AB}} : V_{AB} \to V_{AB}$ is likewise a linear operator on a finite-dimensional complex vector space other than $\left\{ 0 \right\}$, for the same reason as in Part 1 it has an eigenvalue $\lambda_{C} \in \mathbb{C}$ and a corresponding eigenvector $v \ne 0$. Since $v \in V_{AB} \subset V_{A}$, all of the following hold.

$$ Av = \lambda_{A}v, \quad Bv = \lambda_{B}v, \quad Cv = \lambda_{C}v $$

That is, the nonzero vector $v$ is a simultaneous eigenvector for $\mathcal{A} = \left\{ A, B, C \right\}$, and $\lambda_{A}$, $\lambda_{B}$, $\lambda_{C}$ are the simultaneous eigenvalues corresponding to $v$.

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