So what's the point of linear algebra, anyway?

4 points by 3PS 1 year ago | 5 comments
  • richrichie 1 year ago
    People have written very nice books on this e.g.

    https://linear.axler.net/

    • 3PS 1 year ago
      > The novel approach taken here banishes determinants to the end of the book.

      Big fan of this approach! Though I have warmed up to determinants ever since I saw 3Blue1Brown give a fairly intuitive explanation for them [0].

      I'm kind of curious as to how they covered eigenvalues/the characteristic polynomial without determinants. Maybe they just jumped straight to diagonalization?

      [0] https://www.youtube.com/watch?v=Ip3X9LOh2dk

      • richrichie 1 year ago
        One does not need determinant to define eigenvalues. For example:

        If T is a linear operator on vector space V, a scalar a is an eigenvalue if there is a v in V s.t. Tv = av.

        This is the approach the book takes.

        • 3PS 1 year ago
          I agree, but the definition alone isn't sufficient to actually calculate eigenvalues. Hence the standard approach which says that for matrix A, vector v, and eigenvalue λ, we have

            Av = λv
  => Av - λv = 0
  => (A - λI)v = 0
  => det(A - λI) = 0
          Which then yields the characteristic polynomial. Skipping the determinant means you need a different approach.