Hilbert Spaces and Orthogonal Projection in Infinite Dimensions · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.
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Projecting in infinite dimensions

Mathematics · Calculus & Analysis · ages 23-24
Name ______________________   Date ____________
  1. In real R^2 with the usual dot product, project (7,1) onto the line spanned by (1,0). What is the projection's first coordinate?

    Answer: ______________

  2. What does completeness require of an inner-product space with its induced norm?

    • Every linear subspace contains all limits of its convergent sequences
    • Every Cauchy sequence converges to a vector in the space
    • Every orthonormal family has only finitely many unit vectors
  3. In a Hilbert space, partial sums of an orthonormal series with square-summable coefficients converge in norm to a vector in the space.

    Circle one:   True   False

  4. Let p belong to a closed subspace M of a Hilbert space. For p to be the projection of v, what must r=v-p satisfy?

    • It must equal v even when p is a nonzero vector
    • It must lie in M and equal the candidate vector p
    • It must be orthogonal to every vector in M
  5. A vector has squared norm 25. In an orthonormal basis its only nonzero coefficients are 3 and c. What is |c|?

    Answer: ______________

  6. Why does the Hilbert-space closest-point theorem require the target subspace to be closed?

    • Finite orthonormal sums would otherwise stop satisfying Parseval's identity
    • The limit of a minimizing sequence might otherwise lie outside the subspace
    • All bounded linear functionals would otherwise become unbounded on the ambient space
  7. In an orthonormal basis, a vector's only nonzero coefficients are 6 and 8. What is its squared norm?

    Answer: ______________

  8. For v=(2,3) and M the x-axis, Sam proposes p=(2,1). The residual (0,2) is orthogonal to M. Is p the projection?

    • No, because p does not belong to M
    • Yes, because orthogonality alone determines the projection
    • Yes, because p and v have the same first coordinate
    • No, because an orthogonal residual must belong to M
  9. In c00 with the l2 norm, let e_n be the standard coordinate vectors. Leo forms s_N=sum_{n<=N}(2/n)e_n. Why is there no norm limit in c00?

    • An inner-product space cannot contain an infinite orthonormal sequence
    • The coefficient sequence has an infinite sum of squared absolute values
    • The limit has infinite support, and c00 is not complete
  10. On a complex Hilbert space with inner product linear in the first argument, f(x)=<x,y>. The functional i times f is represented by -i times y.

    Circle one:   True   False

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Answer key

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Projecting in infinite dimensions W1-mt_xmKq8SAgvy-s1

  1. 7 · The dot-product ratio is 7/1. Multiplying (1,0) by 7 gives projection (7,0).
  2. Every Cauchy sequence converges to a vector in the space · Completeness means that every norm-Cauchy sequence has a norm limit inside the space.
  3. True · Squared tail norms are tails of the convergent sum of squared coefficients. The partial sums are Cauchy, and completeness supplies their norm limit.
  4. It must be orthogonal to every vector in M · The prompt states that p belongs to M. The remaining condition is that v-p be orthogonal to M.
  5. 4 · Parseval gives 25=9+|c|^2. Thus |c|^2=16 and |c|=4.
  6. The limit of a minimizing sequence might otherwise lie outside the subspace · The minimizing sequence is Cauchy. Completeness gives its ambient limit, but closedness is what puts that limit in the subspace.
  7. 100 · Parseval gives 6^2+8^2=36+64=100. All other coefficients contribute zero.
  8. No, because p does not belong to M · The residual condition holds, but p has second coordinate 1 and is not on the x-axis. Both projection conditions are required.
  9. The limit has infinite support, and c00 is not complete · The coefficients are square-summable, so the sums are Cauchy. Their l2 limit is (2,1,2/3,...), which is not finitely supported.
  10. True · The second argument is conjugate-linear: <x,-iy>=i<x,y>=if(x).
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