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Matrix equation

  1. Dec 26, 2007 #1
    We have [tex]X\in R^{s\times n}[/tex], [tex]A\in R^{n\times s}[/tex], [tex]I\in R^{s\times s}[/tex], where [tex]I[/tex] stands for the indentity matrix.
    Now if we assume that [tex]rank(A)=s[/tex], can we conclude that there must exist a solution [tex]X[/tex] to matrix equation [tex]XA=I[/tex]?

    For me the answer is obviously "yes" if we think this problem in the language of linear map instead of matrices.

    Let [tex]T[/tex] be the linear map whose matrix is [tex]A[/tex] with respect to the standard basis. Then the above question is converted into if [tex]T[/tex] is injective, can we always find an [tex]S[/tex] maping [tex]R^{n}[/tex] to [tex]R^{s}[/tex], such that [tex]ST[/tex] is an identity operator in [tex]R^{s}[/tex]?

    The linear map [tex]S[/tex] is defined as follows:
    for any [tex]x\in range(T)[/tex], [tex]Sx[/tex] is the unique vector [tex]y\in R^{s}[/tex] such that [tex]Ty=x[/tex]. this is ensured because [tex]T[/tex] is injective.
    for other [tex]x\in R^{n}[/tex], [tex]Sx=0[/tex].

    Though knowing for sure how [tex]S[/tex] behaves, is there any way to write out the matrix of [tex]S[/tex] with respect to the standard basis?
    i.e. what does the solution [tex]X[/tex] to matrix equation [tex]XA=I[/tex] look like?
  2. jcsd
  3. Dec 28, 2007 #2
    Use singular value decomposition,

    [tex]A=U\Sigma V^T \Longrightarrow X = V\hat{\Sigma}U^T[/tex]

    where [tex]\hat{\Sigma}[/tex] has the reciprocals of the original on the diagonal. Note that there is a technical peculiarity but I guess you can figure it out. Search for left inverse (or pseudo-inverseor Moore-Penrose inverse). It is almost fun .
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