CounterexampleConstruction: AI-generatedVerification: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30

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An invariant subspace need not reduce an operator

Statement

On $C^{2}$ with its standard inner product, let $T$ have matrix $T = (\begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array})$ in the standard orthonormal basis $e_{1}, e_{2}$ , and let $M = span {e_{1}}$ . Using the convention that $M$ reduces $T$ when both $M$ and $M^{⊥}$ are $T$ -invariant, $M$ is invariant but does not reduce $T$ : its orthogonal complement $M^{⊥} = span {e_{2}}$ is not $T$ -invariant. If $Q$ is the orthogonal projection onto $M^{⊥}$ , then $Q T Q = 0$ even though $T e_{2} = e_{1} \neq 0$ .

Facts & Assumptions

Given: The standard orthonormal basis $e_{1}, e_{2}$ of $C^{2}$ , the displayed linear operator $T$ , and $M = span {e_{1}}$ .

[A1]

A subspace $W$ is $T$ -invariant exactly when $T (W) \subseteq W$ (Invariant subspaces, restrictions, and induced quotient operators).

[A2]

A vector is in $S^{⊥}$ exactly when it is orthogonal to every vector of $S$ (Orthogonality and the orthogonal complement).

[A3]

For a finite-dimensional inner-product space and subspace $W$ , the orthogonal projection $P_{W} x$ is the unique $W$ -component of $x$ in $V = W \oplus W^{⊥}$ (The orthogonal projection $P_{W} v$ is the $W$ -component in $V = W \oplus W^{⊥}$ ).

Proof

technique · direct

Given: The data in the statement and facts [A1]–[A3].

1.1algebra

Matrix multiplication gives $T e_{1} = e_{1}$ and $T e_{2} = e_{1}$ , so for all $a, b \in C$ , $T (a e_{1} + b e_{2}) = (a + b) e_{1}$ .

1.2A2A3

For $x = a e_{1} + b e_{2}$ , $b e_{2} \in M^{⊥}$ and $a e_{1} \in (M^{⊥})^{⊥}$ , so the orthogonal decomposition in [A3] gives $Q x = P_{M^{⊥}} x = b e_{2}$ .

2.1A1A2step 1.1

Every $m = a e_{1} \in M$ satisfies $T m = a e_{1} \in M$ , so $T (M) \subseteq M$ and $M$ is invariant by [A1]. Since $e_{2} \in M^{⊥}$ but $T e_{2} = e_{1} \notin M^{⊥}$ , we also have $T (M^{⊥}) ⊈ M^{⊥}$ by [A2] and step 1.1.

3.1step 1.1step 1.2step 2.1∎

For every $x = a e_{1} + b e_{2}$ , step 1.2 and step 1.1 give $Q T Q x = Q T (b e_{2}) = Q (b e_{1}) = 0$ , while step 1.1 gives $T e_{2} = e_{1} \neq 0$ . Thus the orthogonal compression vanishes although $T$ has a nonzero block from $M^{⊥}$ into $M$ ; combining this with step 2.1 proves the stated counterexample.

Depends on

Used by

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Sources

Kostenko, Trace Ideals with Applications, §3.4 (standard reference, not scraped)
van Neerven, Functional Analysis, §14.5.a (standard reference, not scraped)
Dyatlov–Zworski, Mathematical Theory of Scattering Resonances, App. B §§B.5–B.6 (standard reference, not scraped)