Can Injectivity of a Matrix Guarantee a Left Inverse?

[y_lindsay]

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

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

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

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

Though knowing for sure how S behaves, is there any way to write out the matrix of S with respect to the standard basis?
i.e. what does the solution X to matrix equation XA=I look like?

 

[trambolin]

Use singular value decomposition,

A=U\Sigma V^T \Longrightarrow X = V\hat{\Sigma}U^T

where \hat{\Sigma} 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 .