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 .