Orthogonal Complements

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  • Опубликовано: 7 ноя 2024

Комментарии • 12

  • @speedbird7587
    @speedbird7587 2 года назад

    very intuitive and neat explanation, professor.
    indeed, your lectures seem much more intuitive and clear than the Kenneth Hoffman linear algebra book!

  • @michellejingdong
    @michellejingdong Год назад

    Thank you so much for the videos!

  • @mateuszbaginski5075
    @mateuszbaginski5075 Год назад

    What is the complement of the empty set? If it's V, then we have an exception to (U^\bot)^\bot=U because (\emptyset^\bot)^\bot=\{0\}

    • @sheldonaxler5197
      @sheldonaxler5197  Год назад

      The orthogonal complement of the empty set is indeed V. However, the theorem that (U^\bot)^\bot=U has as a hypothesis that U is a subspace of V. The empty set is not a subspace of V and thus is not an exception to this result.

  • @speedbird7587
    @speedbird7587 2 года назад

    Thanks for your brilliant videos, professor. It seems all of the proofs here makes the assumption that the vector space is finite dimensional. Would you give some nonexamples in which the theorems are not true for infinite dimensional vector space?

    • @sheldonaxler5197
      @sheldonaxler5197  2 года назад

      See, for example, Exercise 14 in Section 6.C of the book for an example of a result about orthogonal complements that does not hold in the infinite-dimensional case.

  • @saiouln.8060
    @saiouln.8060 6 лет назад +1

    Excuse me, Sheldon Axler, sir, how do we initially come up with the idea to let the coefficient in the linear combination be the scalar production of v and each component of the basis {e1,e2,....em}?

    • @sheldonaxler5197
      @sheldonaxler5197  6 лет назад

      Your question is answered in the proof of 6.30 in the book. Note that Chapter 6 of the book is freely available on the book's website (linear.axler.net/).

  • @fsaldan1
    @fsaldan1 4 года назад +1

    In the book, in the proof of 6.51, and given 6.52, wouldn't it be easier to prove that dim((Uperp)perp) = dim(U)? Indeed dim((Uperp)perp) + dim(Uperp) = dim(V) and dim(U) + dim(Uperp) = dim(V).

    • @mkkkk1643
      @mkkkk1643 2 года назад

      6.51 doesn’t state v is finite dimensional so u cannot use 6.50.

    • @spiderjerusalem4009
      @spiderjerusalem4009 Год назад

      ​@@mkkkk1643neither holds in infinite dimensional

    • @mkkkk1643
      @mkkkk1643 Год назад

      @spiderjerusalem4009 just simply read the proof