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Part BCSIR NET December 2025orthogonal-to-one-eigenvector-is-a-3-dimensional-space-not-automatically-another-eigenspace

Orthogonal to one eigenvector is a 3 dimensional space not automatically another eigenspace

Let B be a 4×4 positive-definite real symmetric matrix which is not the identity matrix. Consider the inner product on R4\mathbb{R}^{4} given by ⟨v,w⟩ = vᵀBw, where vᵀ denotes the transpose of v. Which of the following statements is FALSE?

  1. A.If v and w are eigenvectors of B for distinct eigenvalues, then ⟨v,w⟩ = 0.
  2. B.If v is an eigenvector of B and w ≠ 0 is such that ⟨w,v⟩ = 0, then w is an eigenvector of B.✓ Correct
  3. C.For every subspace W⊆R4,W+W⊥=R4W \subseteq \mathbb{R}^{4}, W + W\perp = \mathbb{R}^{4}, where W⊥ = {v∈R4v \in \mathbb{R}^{4} ∣ ⟨v,w⟩ = 0 for all w ∈ W}.
  4. D.If W⊆R4W \subseteq \mathbb{R}^{4} is an eigenspace of B, then W has an orthonormal basis.

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests converse assumed.

See pricing

50 are analysed free — try those first.

The trap it tests

Converse assumed

The theorem runs one way. You used it in the other.

Drill statements like this

Related counterexample: A real matrix with all real eigenvalues is orthogonally diagonalisable

More on this topic

The chapter behind this: Inner products, orthogonality and the spectral theorem — free to read

From Inner Product Spaces and Forms › Gram–Schmidt, orthogonal/unitary/normal matrices, spectral theorem

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