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Part CCSIR NET December 2024orthogonal-pair-does-not-make-an-orthonormal-basis

Orthogonal pair does not make an orthonormal basis

Let V be the R−\mathbb{R}-vector space of real valued continuous functions on the interval [0,π][0, \pi] with the inner product given by ⟨f, g⟩ =∫0πf(x)g(x)= \int_{0}^\pi f(x)g(x)dx. Let S = {sin⁡(x),cos⁡(x),sin⁡2(x),cos⁡2(x)\sin(x), \cos(x), \sin^{2}(x), \cos^{2}(x)} and W be the subspace of V generated by S. Which of the following statements are true?

  1. A.S is a basis of W.✓ Correct
  2. B.S is an orthonormal basis of W.
  3. C.There exist f, g ∈ S such that ⟨f, g⟩ = 0.✓ Correct
  4. D.S contains an orthonormal basis of W.

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.

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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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