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Part BCSIR NET December 2024check-membership-before-orthonormality

Check membership before orthonormality

Consider R4\mathbb{R}^{4} with the standard inner product. Let V be the subspace of R4\mathbb{R}^{4} spanned by the vectors (1, 0, 0, 1), (0, 1, 0, 1), and (0, 0, 1, 0). Which of the following is NOT an orthonormal basis of V?

  1. A.{(−1/2,0,0,−1/2),(−1/6,2/3,0,1/6),(0,0,1,0)(-1/\sqrt{2}, 0, 0, -1/\sqrt{2}), (-1/\sqrt{6}, \sqrt{2}/\sqrt{3}, 0, 1/\sqrt{6}), (0, 0, 1, 0)}
  2. B.{(1/2,0,0,1/2),(−1/6,2/3,0,1/6),(0,0,1,0)(1/\sqrt{2}, 0, 0, 1/\sqrt{2}), (-1/\sqrt{6}, \sqrt{2}/\sqrt{3}, 0, 1/\sqrt{6}), (0, 0, 1, 0)}
  3. C.{(1/2,0,0,1/2),(−1/6,2/3,0,−1/6),(0,0,1,0)(1/\sqrt{2}, 0, 0, 1/\sqrt{2}), (-1/\sqrt{6}, \sqrt{2}/\sqrt{3}, 0, -1/\sqrt{6}), (0, 0, 1, 0)}✓ Correct
  4. D.{(1/2,0,0,1/2),(1/6,−2/3,0,−1/6),(0,0,1,0)(1/\sqrt{2}, 0, 0, 1/\sqrt{2}), (1/\sqrt{6}, -\sqrt{2}/\sqrt{3}, 0, -1/\sqrt{6}), (0, 0, 1, 0)}

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 hypothesis dropped.

See pricing

50 are analysed free — try those first.

The trap it tests

Hypothesis dropped

A theorem applied without checking one of its preconditions.

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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