$A$ has more columns than rows and has full row rank, show there exist infinitely many $B$ s.t. $AB=I$

Question

If A $\in M_{m\times n}(R)$ such that $n>m$.

Prove that if $\text{rank} (A) = m$ then there are infinitely many matrices $B \in \ M_{n\times m} (R)$ such that $ AB = I_m$

So the question is defining a matrix $A$, in a set of matrices where the number of columns is always greater than the number of rows, and if $A$ has full row rank, then there are infinitely many matrices $B$ such that $AB$ creates the identity matrix.

I'm not sure where to go with this question, could anyone please help?

ETA

Thank you to everyone for their answers, the question makes sense now.

Answer 1

Full row rank implies existence of at least one right inverse, because the linear system $Ax=b$ has a solution for every $b$; if $b_i$ is a solution of $Ax=e_i$ (where $e_i$ is the $i$-th column of the $m\times m$ identity) and $B=[b_1\ b_2\ \dots\ b_m]$, then $AB=I$ by construction.

Next we show that we can modify $B$ without changing the result. Let $v\ne0$ be any vector such that $Av=0$. Such a vector surely exists because of the rank-nullity theorem (alternatively, because the consistent system $Ax=0$ has free variables). Consider the matrix $$ C=[\,\underbrace{v\ v\ \dots\ v}_{m}\,] $$ so that $$ A(B+C)=AB+AC=AB+0=I_m $$ Since there are infinitely many choices of a vector $v$, you're done.

Answer 2

The matrix $A$ is a surjective linear transformation $\mathbb{R}^n \to \mathbb{R}^m$ and its kernel, say $K$, has dimension $k = n - m > 0$.

Take a standard basis vector $e_i \in \mathbb{R}^m$. Since $A$ is surjective and not injective (kernel has positive dimension) you have infinitely many preimages w.r.t $A$. Hence take some $b_i \in \mathbb{R}^n$ such that $A b_i = e_i$.

Now define $B \colon \mathbb{R}^m \to \mathbb{R}^n$ by simply declaring that the standard basis vector $e_i \in \mathbb{R}^m$ is mapped to $b_i$, i.e. $B(e_i) = b_i$. This amounts to say that the $i$-th column of $B$ is precisely $b_i$.

By construction $A B (e_i) = A b_i = e_i$ for any $i = 1, \dots, m$. Hence $A B = I_m$.

Answer 3

The objective of this answer is to give a quite general form for matrices $B$.

The full rank condition implies that there exist a $m \times m$ invertible submatrix of $A$.

Let us assume at first that the left-most block $C$ of $A$ is such a submatrix.

Then, with the following notations:

$A=\begin{pmatrix} C & D \end{pmatrix} \ \ \ (1)$

$B=\begin{pmatrix} C^{-1}(I_m-DE)\\ E \end{pmatrix} \ \ \ (2) $

we have $AB=I_m$ for any $E$.

Remark:

The set of such matrices $B$ is an affine space $R=v+V$, where $V$ is a vector space with dimension at most $m \times (n-m)$.

If the leftmost block is not invertible, consider the columns $A_{k_1}, A_{k_2}, \cdots A_{k_m}$ of $A$, which assembled, give an invertible matrix. We consider the $n \times n$ permutation matrix that sends $k_1, k_2, \cdots k_m$ to $1,2,\cdots, m$, and leaves unchanged the other indices of $1,2...n$. It is rather easy to see that if the lines of $B$ (as described by (2)) are "scrambled" by the same permutation, the result is preserved.

(one can give a more rigorous treatment for this last part).