Linear Algebra 4h: Unions and Intersections of Linear Subspaces

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  • Опубликовано: 11 дек 2024

Комментарии • 28

  • @MathTheBeautiful
    @MathTheBeautiful  4 года назад +2

    Go to LEM.MA/LA for videos, exercises, and to ask us questions directly.

  • @lukedavis6711
    @lukedavis6711 9 лет назад +17

    this is the best exposition of this topic I've heard. this was really unnecessarily confusing in David C. Lay and Sheldon Axler. thank you for this!

    • @xoppa09
      @xoppa09 6 лет назад +4

      he is a master teacher :)

  • @darrenpeck156
    @darrenpeck156 2 года назад +2

    Please consider covering modules. Would benefit so many people. Thank you!

  • @antonellomascarello4698
    @antonellomascarello4698 2 года назад +2

    6:03 Geometric Vectors
    8:22 The union is not a subspace
    10:17 The intersection is a subspace
    14:02 IMPORTANT NOTE

  • @rickygallonegro2901
    @rickygallonegro2901 3 года назад +2

    I think this is difficult to grasp because we imagine numbers to be super expansive and when we read about the intersection the only example we get is the trivial example. This guy really is a master teacher.

  • @xoppa09
    @xoppa09 6 лет назад +5

    13:20 I am following your argument that the intersection of two subspaces is also a vector space, the vector sum part.
    How is this argument.
    Let A,B be vector subspaces of vector space V.
    Show that v1 + v2 ∈ A ∩ B whenever v1,v2 ∈ A ∩ B.
    Proof:
    1. v1,v2 ∈ A ∩ B implies v1,v2 ∈ A and v1,v2 ∈B.
    2. v1,v2 ∈A implies v1 + v2 ∈A (since we assumed A is a subspace).
    3. v1,v2 ∈B , then v1+ v2 ∈B (since we assumed B is a subspace).
    4. By steps 2 & 3, v1 + v2 ∈A and v1 + v2 ∈B.
    5. By the definition of set intersection, v1 + v2 ∈A ∩ B .

  • @OtterMorrisDance
    @OtterMorrisDance 3 года назад +1

    Another great video, thanks Pavel!

  • @agh1750
    @agh1750 4 года назад +1

    Your explanation made perfect sense without me having to pause the video to "explain it to myself".

    • @MathTheBeautiful
      @MathTheBeautiful  4 года назад +2

      I'm glad that happened, but pausing the video and explaining it to yourself is also a great way to learn!

  • @maxpercer7119
    @maxpercer7119 2 года назад +1

    pavel i am going to watch all your videos. this is it

  • @xoppa09
    @xoppa09 7 лет назад +3

    Just a clarification, the way you defined vectors the zero vector is always at the origin (or a dot at the origin). That way vectors are always emanating from the origin. Or to put it another way, if you shrink vectors they will vanish at the origin.

  • @boutiquemaths
    @boutiquemaths 2 года назад +1

    Exciting end to the section! The playlist is missing a section 5? Will we be skipping any content if we start watching the videos in section 6?

  • @MarkLeavenworth
    @MarkLeavenworth 5 лет назад +1

    This actually took me nearly an hour to explain to myself, but this thought is what helped me: On union: 'Is it possible for that which admits of only one linear subspace to be in the same linear subspace with that which admits of only being in another linear subspace?...and on intersection: 'Is it possible for that which admits of having two linear subspace properties, or belonging in two linear sub spaces simultaneously to not exist in any linear subspace?'

  • @manishbhatta4391
    @manishbhatta4391 7 лет назад +1

    Good Explaination . Prima!!!!

  • @Sara-cq6gt
    @Sara-cq6gt 8 месяцев назад

    great explanation. thanks!

  • @samchan2535
    @samchan2535 8 лет назад +1

    Insightful.

  • @alekssandroassisbarbosa3749
    @alekssandroassisbarbosa3749 7 лет назад

    is possible to see the exercises before the video solution?

  • @kunal29041
    @kunal29041 7 лет назад

    In the geometric vector space, consider a plane parallel to x-y plane as a subspace, and y-z plane as another subspace, their intersection will be a line. My question is if we select any vector from that line, multiplied by 0, we will get zero vector, which does not belong to that subspace. So, does that mean intersection of two subspaces is not linear subspace?

    • @roodra1154
      @roodra1154 7 лет назад +1

      I think there is a flaw in your logic. The zero vector does belong to that subspace! I think the confusion lies in your understanding of the geometric zero vector. The zero vector is just a line who's starting point and endpoint are the same. This means that for geometric vectors the zero vector is just a point! It's defined this way because V+0 must equal V. So the only geometric structure that one can add to a line and get the same line is a point! And all geometric vectors have the zero vector because you can just take a point from the line/plane. I hope this helps clear up some confusion.

    • @xoppa09
      @xoppa09 6 лет назад +5

      The problem with your logic is "consider a plane parallel to x-y plane as a subspace". A plane parallel to the x-y plane is not a subspace. Think about it, a plane parallel to the x-y plane does not contain the zero vector.

    • @alojzybabel4153
      @alojzybabel4153 4 года назад +1

      *spring1* is right, your mistake is in assuming that the plane parallel to the XY plane is a subspace, which it isn't, because it does not contain the 0 vector (containing the 0 vector = the origin is the dead give away to tell if something is a subspace or not, because if it doesn't contain the 0 vector, it cannot possibly be a subspace). That parallel plane is a subSET, but it is not a subSPACE.

  • @ImprEcr
    @ImprEcr 5 лет назад

    Ok ... I'm on point . Thanks by the way, beautiful.

  • @aeshray9003
    @aeshray9003 8 лет назад +1

    helpful indeed