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I know that a question that is very similar, other than the wording, was asked here, but I am more interested in being critiqued on my proof. I would like to know if it is correct, if it is complete, and if it is correctly formatted. Instead of just writing out the answer in a linear fashion and moving on to the next problem, I want to use this opportunity to practice my proof writing skills.

I would also like to know if I should include this lemma: The inverse of $AB$ is the reverse product $B^{−1}A^{−1}$. If it should be included, where does it belong in the proof? Before I even state the theorem? As one of the steps?

Theorem: if $B$ is the inverse of $A^2$, then $AB$ is the inverse of $A$.

Proof: Assume $B$ is the inverse of $A^2$.

1) $A^{ 2 }=AA$, so $[A^{ 2 }]^{ -1 }=(AA)^{ -1 }$

2) The inverse of $AA$ is the reverse product $A^{ -1 }A^{ -1 }$

3) Therefore, $B=A^{ -1 }A^{ -1 }$

4) $(A)B = (A)A^{ -1 }A^{ -1 }$

5) $AB = IA^{ -1 }$

6) $AB = A^{ -1 }$

Q.E.D.

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Cherry_Developer
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  • That proof looks good. For the fact that $(AB)^{-1}=B^{-1}A^{-1}$ you could prove that as a lemma before the theorem, although that just seems like a fact about inverses that is assumed to be known. However, if this proof is for a homework assignment where that fact has not been discussed in class I would include a proof for it as a lemma to be safe. Also, it seems like $BA$ could also be the inverse of $A$ (that is $BA=AB$). – Dave Apr 30 '16 at 17:25
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    You are implicitly using that $A$ is invertible. You should explicitly justify that inference. More simply $\ A(AB) = A^2B = 1,,$ and $\ (AB)A = (BA^2)(AB)A = BA(AAB)A = BA^2 = 1,$ by associativity and hypotheses. – Bill Dubuque Apr 30 '16 at 17:32
  • @BillDubuque So, the first few steps of my proof have to demonstrate that $A$ is invertible and then I can carry on with the steps that I already wrote up? – Cherry_Developer Apr 30 '16 at 18:00
  • By assuming $A^2$ is invertible, we can see that $A$ is invertible by the multiplicative property of determinants. – Dave Apr 30 '16 at 19:29

1 Answers1

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Hint:

$B$ is the inverse of $A^2$ means:

$$ A^2B=I \iff A(AB)=I $$

so $AB$ is the inverse of $A$

Emilio Novati
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  • I'm confused as to how this is just a hint. It seems that you proved the implication with just $A^2B=I \iff A(AB)=I$. Is there anything else that needs to be done? – Cherry_Developer Apr 30 '16 at 17:56
  • Yeah.. it is more than an hint....You have also to add that a right inverse is also a left inverse and remember that the inverse is unique. – Emilio Novati Apr 30 '16 at 18:11
  • I just checked the answer in the back of the book I am using and it seems to have given the answer that you already wrote up. This is what it saids: $A^{ 2 }B=I$ can also be written as $A(AB)=I$, therefore ${ A }^{ -1 }$ is $AB$. Why is the answer the author used not as rigorous as what you are suggesting? I would also like to know if you would be willing to guide me right now in formulating the correct and complete proof for this. I am new to proof writing. – Cherry_Developer Apr 30 '16 at 19:49
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    Depending on what you want be ''pedantic'' in your proof you can consider the statement in my answer as complete or not. In a ''pedantic'' approach you have to cite the fact that the ''iff' from $A^2B=I$ and $A(AB)=I$ is a consequence of associativity and the fact that the proof is write also for a right inverse is not relevant because the inverse is unique. – Emilio Novati Apr 30 '16 at 20:28
  • Ok, I will reiterate to you what I have understood from your explanations. Let me know if I got it right, please. So, we know that $A^2B=I$ can also be written as $A(AB)=I$, therefore, ${ A }^{ -1 }$ is $AB$. This is done by just using the associative property and it never assumes that $A$ is invertible (like I originally did in step 2 of my proof). It just shows that $A$ is invertible (by AB) because $A$ multiplied by (AB) is the identity. – Cherry_Developer May 01 '16 at 00:03
  • In regards to what you said about having to address the right inverse, we know that a matrix $A$ cannot have two different inverses. Suppose $BA=I$ and also $AC=I$. Then $B=C$. So, $B(AC)=(BA)C$ which gives $BI=IC$ or $B=C$. So we have shown that a left inverse B and a right inverse C must be the same matrix. Is this what you are talking about when you say that the inverse is unique? – Cherry_Developer May 01 '16 at 00:08
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    Yes, it's all right! – Emilio Novati May 01 '16 at 15:51