Two Methods to Find Standard Matrix for Projection Onto a Line [Passing Linear Algebra]

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  • Опубликовано: 15 сен 2024

Комментарии • 13

  • @MissSixlane
    @MissSixlane 5 лет назад +1

    So helpful! Thank you so much!

  • @kamogelomorole256
    @kamogelomorole256 4 года назад +1

    Good explanation 👊🏾

  • @omarshoukry4402
    @omarshoukry4402 4 года назад +1

    Very helpful!!! Thank you

  • @jamiedeal8160
    @jamiedeal8160 4 месяца назад +1

    I'm confused how to apply this when the span of W is more than one vector

    • @crystalfruit1193
      @crystalfruit1193 3 месяца назад

      I think if you use gram-schmidt thats how you solve it. Not sure though

  • @jhonsonholland5093
    @jhonsonholland5093 9 месяцев назад +1

    I was wondering if the second method would work for a subspace more than 1dimension. How would we use the second method if the subspace was 2 dimension in R^3???

    • @Anthony_Delcoro
      @Anthony_Delcoro 9 месяцев назад

      It does, that is exactly what my homework assignment is asking about.

  • @samirkhan6195
    @samirkhan6195 2 года назад

    Another simple method is by recognising the fact that orthogonal projection matrix will be symmetric matrix , let's call this P, and from spectral theorem - every symmetric matrix is orthogonally diagonalizable , and you already have all the required component of this orthogonal diagonalization , We know eigenvalues (1,0,0) , we know eigenvector corresponding to eigenvalue 1 , i.e. above given vector v_1 , Don't forget to normalize this, lets call normalized vector b_1, other eigen vectors are not required because we are gonna use Spectral decomposition and components corresponding to 0 eigenvalues will be anyway 0 ...
    A = 1(b_1 b_1^T) --- That's it.

  • @ronak859
    @ronak859 3 года назад +3

    Does this work in R2 for projecting onto a line

  • @johntapp7980
    @johntapp7980 3 года назад

    Thank you!

  • @Stevez221
    @Stevez221 3 года назад

    Thanks!

  • @aarons8711
    @aarons8711 2 года назад

    🖐🤚