# Proof: $CA = I_n$ and $AD = I_m \Rightarrow C = D$

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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To see how the hint applies, try this:

$C=CI_m=CAD=I_nD=D$

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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