Lecture - 2.1 Linear Independence

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  • Опубликовано: 26 дек 2024

Комментарии • 10

  • @dhivyadharshini9698
    @dhivyadharshini9698 3 года назад +5

    Sir I could easily catch up the notion thanks to your teaching yet it would be more clear if you solve the exercise sums atleast one or two instead of putting everything as exercises.

  • @psh_IITD
    @psh_IITD 4 года назад +3

    Sir teaching is superb, please add some quality examples after every results.

  • @techtangle8046
    @techtangle8046 3 года назад +3

    sir, you are highlighting everything, but an illustration or figure could be more helpfull ??

  • @chiragchatwani9124
    @chiragchatwani9124 4 года назад +3

    I did not unterstood the proof what to do
    please help

  • @epistemophilicmetalhead9454
    @epistemophilicmetalhead9454 10 месяцев назад

    linear dependence: linear combination = 0 despite at least one coefficient being non-zero OR when at least one vector in the linear combination can be expressed as a combination of other vectors in the span
    Null set is linearly independent
    subset of a linearly independent set is also linearly independent.

  • @piyushjoshi5462
    @piyushjoshi5462 3 года назад +2

    Sir if it's true that subspace is always linearly dependent? As it is subset of vector space V and it contains zero vector

    • @praveenbhati3497
      @praveenbhati3497 2 года назад

      i too have same doubt and in this manner space should also be alway linearly dependent cause it too contain 0 vector

    • @skrejuan_
      @skrejuan_ 2 года назад

      Yes any subspace and the whole vector space are linearly dependent

  • @ThakursBoy
    @ThakursBoy Год назад +2

    5done ✓

  • @tikarambhusal8657
    @tikarambhusal8657 2 месяца назад

    S equals S’ then it will be independent