I'd like to compute A ∗ = a r g m a x A ∈ R d × k A T A = I ⁡ det ( A T Λ A )where k ≤ d, Λ = diag ⁡ ( λ 1 , … , λ d ) and λ 1 ≥ ⋯ ≥ λ d ≥ 0.I strongly suspect that A ∗ = [ ϵ 1 ⋯ ϵ k ], but after various attempts to prove it I gave up.

Question

I&#039;d like to compute      A    ∗    =            a      r      g            m      a      x                                                          A              ∈                                                R                                                  d                  ×                  k                                                                                                        A                T                            A              =              I                                            ⁡  det  (      A    T    Λ  A  )where   k  ≤  d,   Λ  =  diag  ⁡  (      λ    1    ,  …  ,      λ    d    ) and       λ    1    ≥  ⋯  ≥      λ    d    ≥  0.I strongly suspect that       A    ∗    =  [      ϵ    1    ⋯      ϵ    k    ], but after various attempts to prove it I gave up.

Turynka2f · Accepted Answer

As   A has orthonormal columns, its singular value decomposition must be in the form of   A  =  U      (                            I                                      0                      )        V    T  . Hence to maximise   det  (      A    T    Λ  A  ) is equivalent to maximise the determinant of       P    k  , the   k  ×  k leading principal submatrix of   P  =      U    T    Λ  U. However, by the interlacing property, the   i-th largest eigenvalue of       P    k   is bounded above by the   i-th largest eigenvalue of   P. Since   P and   U have identical spectra, it follows that   det      P    k    ≤      λ    1    ⋯      λ    k  . Therefore   U  =  I is a maximiser and in turn, every   A  =      (                            I                                      0                      )        V    T   is a maximiser whenever   V  ∈      O    k    (      R    ).

I'd like to compute A ∗ </msup> = <munder> <mrow cl

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