Exercise 2.2.5

Answers

  • 1.
    If A has a rank r = n, then there are r nonzeros in Σ, these are the singular values.
  • 1.
    When A has independent columns, we know that AA is invertible. So we have Σ = UTAV and ΣTΣ = V TATUUTAV = V TATAV , the inverse of ΣTΣ is (ΣTΣ)−1 = V T(ATA)−1V since (ΣTΣ)(ΣTΣ)−1 = V TATAV V T(ATA)−1V = I
  • 1.
    We let Σ+ = (ΣTΣ)−1ΣT = V T(ATA)−1V V TATU = V T(ATA)−1ATU
  • 1.
    Then we have A+ = (ATA)−1AT = (V Σ2V T)−1V ΣUT = V Σ−2V TV ΣUT = V Σ−1UT
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2020-03-20 00:00
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