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In my school book there's an exercise:

We have three matrices $A \in \mathbb{R}^{m \times n}$ and $C, D \in \mathbb{R}^{n \times m}$.
If $CA = I$ and $AD = I$ than we can say that $C = D$.
Proof this.

Hint: work with the product $C\cdot A\cdot D$.

I ignored the hint and I tried to prove it. This is what I got:

\begin{align*} CA = I &\wedge AD = I\\ &\Updownarrow\\ C = A^{-1} &\wedge D = A^{-1}\\ &\Updownarrow \text{ The inverse of a regular matrix is unique}\\ C &= A \end{align*} EDIT: I just saw this proof is wrong, because were not working with square matrices.
Now I've got two questions.

  • Is my own proof correct? I guess not
  • How would you prove it by working with the product $C\cdot A\cdot D$?
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up vote 2 down vote accepted

To see how the hint applies, try this:

$C=CI_m=CAD=I_nD=D$

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1  
QED. There's nothing left to try! – Mark Dec 17 '11 at 16:37
    
@MarkSchwarzmann truth. – user21385 Dec 17 '11 at 16:43
    
@AMPerrine thank you anyway. If I see it, it looks easy. – user21385 Dec 17 '11 at 16:44
    
Sorry if I showed too much. I tried only giving away a part, but since the symmetry of the equations made the other steps very obvious it felt more incomplete than a useful hint. – AMPerrine Dec 17 '11 at 16:50
    
A hint could have been $C(AD)=(CA)D$. – Jeppe Stig Nielsen Sep 7 '15 at 21:36

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