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Part BCSIR NET June 2025the-sin-squared-weight-and-the-2-over-pi-make-these-orthonormal-not-merely-orthogonal

The sin squared weight and the 2 over pi make these orthonormal not merely orthogonal

Let C[0,π]C[0, \pi] be the real vector space of real-valued continuous functions on the closed interval [0,π][0, \pi]. For positive integers n, define fn∈C[0,π]f_{n} \in C[0, \pi] by fn(x)=sin⁡(f_{n}(x) = \sin(nx)/sin x if x∈(0,π),fn(0)=nx \in (0, \pi), f_{n}(0) = n, and fn(π)=(−1)nf_{n}(\pi) = (-1)^{n}⁻1n^{1}n. Let V be the real subspace of C[0,π]C[0, \pi] spanned by {f1,f2,f3f_{1}, f_{2}, f_{3}}. Consider the inner product on V given by ⟨f, g⟩ =(2/π)∫0πf(x)g(x)sin⁡2x= (2/\pi)\int_{0}^\pi f(x)g(x)\sin^{2}x dx. Which of the following statements is true?

  1. A.f4∈Vf_{4} \in V
  2. B.{f1,f2,f3f_{1}, f_{2}, f_{3}} is an orthonormal basis of V.✓ Correct
  3. C.The dimension of V is 2.
  4. D.{f1,f2,f3f_{1}, f_{2}, f_{3}} is an orthogonal set but not orthonormal.

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 standard counterexample.

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50 are analysed free — try those first.

The trap it tests

Standard counterexample

There is a canonical object that settles this. Recognising it is the whole question.

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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