# Proving m=n in matrix problem

• Feb 4th 2011, 08:05 AM
fizzle45
Proving m=n in matrix problem
OK here is the problem
Suppose $\mathbf{A}\text{ is an }m\times n\text{ matrix and there exist matrices }\mathbf{C}\text{ and }\mathbf{D}\text{ such that }\mathbf{CA}=\mathbf{I}_n\text{ and }\\\mathbf{AD}=\mathbf{I}_m\text{. Prove that }m=n$

im soo clueless =( any help would be awesome
• Feb 4th 2011, 02:39 PM
topsquark
Here's a start. Let C be a p x m matrix. Then CA is a (p x m)(m x n) = p x n matrix. But CA = In. Thus C is an n x m matrix. Do this again for AD. What does all this mean for A?

-Dan

I fixed the problem. Unfortunately that makes my comment pointless.

-Dan
• Feb 4th 2011, 03:16 PM
Ackbeet

Does that really work? I can see how you could get that C must be n x m, just to make the sizes work out. Why must it be n x n?

I think you're going to have to use rank arguments here.
• Feb 4th 2011, 03:26 PM
topsquark
You are right. I had a small typo on the worksheet I scratched out the solution on. Apologies and thanks for the catch Ackbeet. 8]

-Dan
• Feb 4th 2011, 04:46 PM
Ackbeet
The following facts may be useful:

$\text{rank}(I_{n})=n,$ and $\text{rank}(I_{m})=m.$

Also, if $A$ is an $m\times n$ matrix, it is true that

$\text{rank}(A)\le\min(m,n).$

Furthermore, it is also true that

$\text{rank}(AB)\le\min(\text{rank}(A),\text{rank}( B)).$

Using these facts, as well as facts about the minimum function, you should be able to prove both that $m\le n,$ and that $n\le m,$ thus proving your result.

Can you see your way forward?