# Show that a matrix has a right inverse

1. Apr 2, 2017

### Mr Davis 97

1. The problem statement, all variables and given/known data
Let $A$ be an $m \times n$ matrix with rank $m$. Prove that there exists an $n \times m$ matrix $B$ such that $AB= I_m$

2. Relevant equations

3. The attempt at a solution
So here is how far I get. I am given that $A$ has rank $m$. Since $L_A(x) = Ax$ is a map $\mathbb{R}^n \rightarrow \mathbb{R}^m$, this means that $L_A$ is a surjective map.
I know that in set theory, surjective maps must have right-inverses, so I get the sense that I am on the right track. But I am not sure how to continue the proof.

2. Apr 3, 2017

### nuuskur

You should be more precise in your expression. The fact that you have a map $\mathbb R^n\to\mathbb R^m$ doesn't imply it's surjective.

What do we know about $m, n$? Right iverses certainly exist [in fact, there's a whole family of such right inverses] when there are more columns than rows and the matrix $A$ has a full row rank.

The Rank-Nullity theorem provides a solution.

3. Apr 3, 2017

### Mr Davis 97

The Rank-Nullity theorem seems to tell me that $\text{nullity}(A) = n-m>0$. But I don't see how this helps me

4. Apr 3, 2017

### StoneTemplePython

my general approach is to make use of orthogonality for these things -- it is a much cleaner cut than general linear independence. Since we are in reals, consider using gram-schmidt and get:

$\mathbf A = \mathbf{T V}^T$

(this is an atypical factorization but one I used in another of your threads recently), or more typically:

$\mathbf A^T = \mathbf{Q R}$, hence $\mathbf A = \Big(\mathbf R^T \mathbf Q^T \Big)$.

Where $\mathbf Q$ is orthogonal, but not full rank (it is n x m) and $\mathbf R$ is square m x m matrix.

What happens if you multiply

$\Big( \mathbf R^T \mathbf Q^T\Big) \Big(\mathbf Q\Big) = \mathbf R^T \big(\mathbf Q^T \mathbf Q\big) = \mathbf R^T$. So we need to be able to invert $\mathbf R^T$... how do I know that $\mathbf R^T$ is invertible?

You could also use Singular Value Decomposition to get this.