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A nonreal eigenvector yields an invariant real two-plane and the standard rotation-scaling block

Statement

Let $T : V \to V$ be an endomorphism of a real vector space, identify $V$ with the fixed real form of the canonical conjugation on $V_{C}$ , and suppose $w = u + i v$ with $u, v \in V$ is an eigenvector of $T_{C}$ with eigenvalue $λ = a + b i$ , where $b \neq 0$ . Then $u$ and $v$ are $R$ -linearly independent, ${span}_{R} {u, v}$ is $T$ -invariant, and with respect to the ordered basis $(u, - v)$ the matrix of $T$ restricted to that plane is

$(\begin{array}{cc} a & - b \\ b & a \end{array}) .$

Moreover $σ (w) = u - i v$ is an eigenvector of $T_{C}$ with eigenvalue $\overline{λ} = a - b i$ .

Facts & Assumptions

Given: A real vector space $V$ , an endomorphism $T$ , and an eigenvector $w = u + i v$ of $T_{C}$ with eigenvalue $λ = a + b i$ , where $u, v \in V$ and $b \neq 0$ .

[L1]

Every element of $V_{C}$ is uniquely $u + i v$ with $u, v$ in the fixed real form $V$ (The fixed points of a conjugation form a real vector space whose complexification recovers the ambient complex space).

[L2]

The canonical conjugation interchanges the generalised eigenspaces of $λ$ and $\overline{λ}$ (For a real operator, nonreal generalised eigenspaces of the complexification occur in conjugate pairs).

Proof

technique · direct

1.1L1givenalgebra

By [L1], the decomposition $w = u + i v$ with $u, v \in V$ is unique. Both $u$ and $v$ are nonzero: if $v = 0$ then $w$ lies in $V$ and $T_{C} w = λ w$ forces $b = 0$ ; symmetrically $u = 0$ would give the same contradiction for the coefficient of $i$ .

2.1step 1.1algebra

Expanding $T_{C} (u + i v) = λ (u + i v)$ gives $T u + i T v = (a u - b v) + i (b u + a v)$ ; comparing the $V$ - and $i V$ -components, which are unique by [L1], yields $T u = a u - b v$ and $T v = b u + a v$ .

2.2step 1.1L2

The conjugate vector is an eigenvector for $\overline{λ}$ : $σ (w) = u - i v$ lies in $G_{\overline{λ}}^{(1)} (T_{C})$ by [L2] applied with $e = 1$ .

3.1step 1.1step 2.1L1algebra

The vectors $u, v$ are $R$ -linearly independent: if $u$ and $v$ were dependent, then, since both are nonzero, $v = α u$ for a real $α$ , so $w = (1 + i α) u$ and dividing the eigen-equation by $1 + i α \neq 0$ gives $T_{C} u = λ u$ . But $T_{C} u = T u$ lies in $V$ , while $λ u = a u + b i u$ has nonzero $i$ -component $b u$ , contradicting uniqueness of the components in [L1].

3.2step 2.1algebra

The plane is invariant and the block appears: $T u = a u - b v = a u + b (- v)$ and $T (- v) = - (b u + a v) = - b u + a (- v)$ by step 2.1, so in the ordered basis $(u, - v)$ the two columns are $(\binom{a}{b})$ and $(\binom{- b}{a})$ .

4.1step 2.2step 3.1step 3.2∎

Steps 3.1 and 3.2 prove the independence, invariance and matrix claims, and step 2.2 the conjugate eigenvector claim.

Depends on

Used by

One nonreal eigenvector reconstructs the invariant real plane of a rotation-scaling block Example

Dependency tree · two levels

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Sources

Keith Conrad, Complexification (notes) (standard reference, not scraped)