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.(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Matrix equation

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