Schur invariant flag

Let A be an n × n matrix with complex entries. If n ≥ 4, which one of the following statements is true?

A.A does not have any non-zero invariant subspace in $\mathbb{C}^{n}$ .
B.A has an invariant subspace in $\mathbb{C}^{n}$ of dimension n − 3.✓ Correct
C.All eigenvalues of A are real.
D. $A^{2}$ does not have any invariant subspace in $\mathbb{C}^{n}$ of dimension n − 1.

The idea, in plain words

Over the complex numbers every square matrix can be tilted into a triangular shape, and once it is triangular you can read off a nested chain of subspaces that the matrix preserves, one of every possible size. Over the reals that fails, which is what several of the options are quietly relying on.

Solution

Over $\mathbb{C}$ every matrix is triangularisable (Schur)

so the span of the first k basis vectors of a triangularising basis is an invariant subspace of every dimension k — in particular n − 3 ≥ 1.

Why each option is right or wrong

Checked against 2 symbolic computations

A.WrongBase field or ring
False over $\mathbb{C}$ , and this is where the field matters.

$\mathbb{C}$ is algebraically closed

so the characteristic polynomial has a root, A has an eigenvector, and its span is a 1-dimensional invariant subspace.

Over $\mathbb{R}$ the statement would be much closer to true — SymPy gives the rotation [[0,−1],[1,0]] eigenvalues ±i

so it has no real eigenvector and no 1-dimensional real invariant subspace.

Read the field before reaching for a counterexample.
B.Correct
True.

By Schur's theorem every complex matrix is unitarily triangularisable: there is a basis in which A is upper triangular.

The span of the first k basis vectors is then invariant for every k

so A has an invariant subspace of each dimension 0, 1, …, n.

Since n ≥ 4, the dimension n − 3 is at least 1 and is one of them.

The trigger: 'complex entries' plus 'invariant subspace' should bring Schur to mind immediately.
C.WrongBase field or ring
False.

Nothing forces the eigenvalues of a complex matrix to be real — SymPy gives iI $_{2}$ the single eigenvalue i with multiplicity 2.

Real eigenvalues are guaranteed for real symmetric matrices, and that hypothesis is absent here.
D.Wrong
False.

$A^{2}$ is itself a complex n × n matrix

so the Schur argument in (B) applies to it verbatim and gives invariant subspaces of every dimension, n − 1 included.

Nothing about squaring destroys the triangularisation.

2 of these 3 wrong options correspond to a named reasoning error — the same errors recur across subjects.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Last revised 12 Sept 2026. Found a mistake? Tell us — corrections are the fastest thing we act on.

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