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I Multiplication by a matrix in GL rotates a plane's basis?

  1. May 6, 2016 #1
    Let [itex] A = (a_{ij}) [/itex] be a [itex] k\times n[/itex] matrix of rank [itex] k [/itex].
    The [itex] k [/itex] row vectors, [itex] a_i [/itex] are linearly independent and span a [itex]k[/itex]-dimensional plane in [itex] \mathbb{R}^n [/itex].

    In "Geometry, Topology, and Physics" (Ex 5.5 about the Grassmann manifold), the author states that for a matrix [itex] g\in \textrm{GL}(k,\mathbb{R}) [/itex],
    [itex] \overline{A} = gA [/itex] defines the same plane as [itex] A [/itex] because [itex] g [/itex] simply rotates the basis within the [itex] k [/itex]-plane.

    I'm having trouble seeing this.
     
    Last edited: May 6, 2016
  2. jcsd
  3. May 6, 2016 #2

    andrewkirk

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    Let ##r_j## denote the ##j##th row of ##A##, and consider a vector ##v## that is normal to the ##k##-dimensional plane. Note that ##v## must be perpendicular to all ##r_j##, so ##r_j\cdot n=0##.

    Then ##Av=0## since the ##j##th component of ##An## is ##r_j\cdot v##.

    So ##\bar Av=(gA)v=g(Av)=g\mathbf 0=\mathbf0##. So ##v## is also normal to the plane defined by ##\bar A##. Since that holds for all ##(n-k)## basis vectors of the null space of ##A##, and the rank of ##\bar A## is the same as that of ##A##, the plane (its rowspace) must be the same.
     
  4. May 7, 2016 #3
    Thank you, I understand now. When you say
    this means that all the ## v ## perpendicular to the ## k##-dimensional plane satisfy ## \overline{A} v = 0##.
     
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