The existence of a matrix? Suppose that the matrix $A\in{\mathbb{R}}^{n\times r}$, $\textrm{rank}(A)=r$, and $I_{n}$ is the identity matrix. Is there a matrix $B\in{\mathbb{R}}^{r\times n}$, such that 
$$AB=I_{n}?$$
What is the requirement for the matrix $A$? Thank you!
 A: From $n=rank (AB)\leq\min\{rank (A),rank (B)\}$ and also $rank(A)\leq\min\{n,r\}$ you can deduce something about the size of $A$ and its rank. 
A: Consider
$$
A=\begin{bmatrix}1\\0\end{bmatrix}
$$
then
$$
AB=\begin{bmatrix}1\\0\end{bmatrix}\begin{bmatrix}b_1 & b_2\end{bmatrix}=\begin{bmatrix}b_1 & b_2\\0 & 0\end{bmatrix}.
$$
Do you see the trouble?
A: One of the solution for $B$ is the Moore-Penrose inverse $A^+$ of $A$.
Solution(s) exist as soon as $n \le r$ and that $rank(A)=n$.
If $n = r$, $B$ is the inverse of $A$.
If $n < r$, the set of solutions corresponds to the Moore-Penrose inverse plus an element ($C$ such that $AC = 0$) owing to a set of dimension equal to $r-n$.
Look at the corresponding Wikipendia entry for example. 
A: Yet another sufficient way of doing this. Since $rank(A) = r$, we have $r\leq n$. Thinking in terms of columns of $A$ will help you see this. You picked $r$ vectors in a $n$-dimensional space. Since $rank(A)=r$, they are all linearly independent. Since you can have only $n$ independent vectors at most in a $n$-dimensional space, $r\leq n$. In case $r=n$, then $B=A^{-1}$. In case $r<n$, $B$ doesn't exist.To see this, observe that determinant of RHS is $1$ ($\neq 0$). Now determinant of LHS (i.e. AB) is zero as it is not a full-rank matrix. To see that, you can think of columns of $AB$ as a linear combination of columns $A$. Since you took $r$ independent vectors and constructed $n(>r)$ vectors out of it, they have to be dependent and hence $AB$ should have determinant zero. 
