Show that a matrix has a right inverse

In summary, the conversation discusses how to prove the existence of an ##n \times m## matrix ##B## such that ##AB=I_m## when ##A## is an ##m \times n## matrix with rank ##m##. The Rank-Nullity theorem and the use of orthogonality and Singular Value Decomposition are suggested as possible approaches to the proof.
  • #1
Mr Davis 97
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Homework Statement


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##

Homework Equations

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.
 
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  • #2
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
nuuskur said:
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.
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
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.
 

Related to Show that a matrix has a right inverse

What is a right inverse of a matrix?

A right inverse of a matrix is another matrix that, when multiplied on the right of the original matrix, results in the identity matrix. In other words, it "undoes" the original matrix when multiplied on the right.

Why is it important to show that a matrix has a right inverse?

Showing that a matrix has a right inverse is important because it guarantees that the matrix is invertible. This means that the matrix has a unique solution to the equation Ax=b for any given vector b.

How do you prove that a matrix has a right inverse?

To prove that a matrix A has a right inverse, you need to find another matrix B such that AB=I, where I is the identity matrix. This can be done by solving the equation Ax=I for x, which will result in a matrix that satisfies the condition for a right inverse.

Can a matrix have more than one right inverse?

No, a matrix can only have one right inverse. If there were two matrices, B1 and B2, that satisfied the condition AB1=I and AB2=I, then by multiplying both sides by B2, we would get B2=I, which contradicts the assumption that B2 is a right inverse of A.

Can a singular matrix have a right inverse?

No, a singular matrix (one that is not invertible) cannot have a right inverse. This is because a right inverse must "undo" the original matrix, and a singular matrix cannot be undone by any other matrix when multiplied on the right.

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