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I am still learning Linear Algebra at it's basic levels, and I encountered a theorem about invertible matrices that stated that:

If $A$ is an invertible matrix, then for $n=0,1,2,3,..$. $A^n$ is invertible and $(A^n)^{-1} = (A^{-1})^n$.

Now, in attempting to write my proof, I proceeded this way (note that it's not complete):

$$A^n(A^{-1})^n=\prod_{i=1}^nA\prod_{i=1}^nA^{-1}=\prod_{i=1}^n(AA^{-1})=\prod_{i=1}^nI=I$$

Is this line of thinking correct? Well, am just returning to math after a long time of little practice, so I could be wrong.

Based on my comment to Dimitri's answer, would my use of this argument improve my proof?

$$\prod_{i=1}^{n-1}A.(AA^{-1}).\prod_{i=1}^{n-1}=\prod_{i=1}^{n-1}A.(I).\prod_{i=1}^{n-1}=...=A.(AIA^{-1}).A^{-1}=AIA^{-1}=AA^{-1}=I$$

After checking the comments, it seems this last argument gives me a correct proof eventually, and I now see that the problem with my original approach was making the argument that:

$$\prod_{i=1}^nA\prod_{i=1}^nA^{-1}=\prod_{i=1}^n(AA^{-1})$$

Which is not necessarily correct, but like @Srivatsan demonstrates, that approach is not at all wrong since :

Notice that $A$ and $A^{−1}$ commute, so this justifies your proof now

Thanks to everyone for guidance, now I see why collaboration is going to make me love math :D

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    $\begingroup$ I think this is wrong because you are using commutativity. Matrix multiplication is not commutative. However, I think you can adapt your proof so it works. $\endgroup$
    – sxd
    Commented Sep 20, 2011 at 13:19
  • $\begingroup$ It's correct if before writing the equality, you write $AA^{-1}=I=A^{-1}A$. $\endgroup$
    – Davide Giraudo
    Commented Sep 20, 2011 at 13:23
  • $\begingroup$ @mcnemesis Actually, that statement is correct, but you have not justified it properly before using it. What is wrong is this: $A^n B^n = (AB)^n$; this is not true for a general pair of matrices. But since $A$ and $A^{-1}$ commute, $A^n (A^{-1})^n = (A A^{-1})^n$ is perfectly fine, provided of course you justify it somewhere. (Check out my answer as well.) $\endgroup$
    – Srivatsan
    Commented Sep 20, 2011 at 14:10
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    $\begingroup$ @mcnemesis Just one final point. By "not wrong", I meant that the logic in sound. But there are two things you should understand clearly. (1.) As long as you don't say why you are allowed to make that switch, the proof is still incomplete, technically speaking. (2.) What level of rigor and detail is necessary completely depends on the audience. In a paper, the author may be justified in omitting such "small" details. (contd) $\endgroup$
    – Srivatsan
    Commented Sep 20, 2011 at 14:21
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    $\begingroup$ (contd) On the other hand, if a text-book carelessly misses this step in a derivation, then either it might mislead students into thinking that you can always switch matrices (i.e., matrix product is commutative, which it isn't), or if the student is sufficiently attentive, it might confuse her ("Why did the author do this? I don't get this step at all."). Finally, in an exam, if you are skipping such steps, you will most likely lose points even if you think your solution is correct. The point of an exam is to show that you understood the concepts. The moral? "When in doubt, show all steps!" $\endgroup$
    – Srivatsan
    Commented Sep 20, 2011 at 14:24

4 Answers 4

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HINT, observe that: $$A^n(A^{-1})^n = \underbrace{A\ldots AA}_{\textrm{n times A}} A^{-1}A^{-1}\ldots A^{-1} = A\ldots A(AA^{-1})A^{-1}\ldots A^{-1} = A\ldots AIA^{-1}\ldots A^{-1}$$

Sorry for the dots, but I didn't find a better way to point the idea out!

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  • $\begingroup$ This is the idea I had in mind, but thought I could condense it by using the $\prod$ operator. And then also, I was assuming that I could expand if necessary to something like: $\prod_{i=1}^{n-1}A.(AA^{-1}).\prod_{i=1}^{n-1}=\prod_{i=1}^{n-1}A.(I).\prod_{i=1}^{n-1}=...=A.(AIA^{-1}).A^{-1}=AIA^{-1}=AA^{-1}=I$. $\endgroup$
    – JWL
    Commented Sep 20, 2011 at 13:45
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    $\begingroup$ That method is indeed correct, here you avoid the error you made in your original proof. The problem with your original proof was that you wrote that $\Pi^n_{i=1} A \Pi^n_{i=1} A^{-1} = \Pi^n_i AA^{-1}$, which uses commutativity. $\endgroup$
    – sxd
    Commented Sep 20, 2011 at 13:54
  • $\begingroup$ Thanks, this clarifies the problem, and your approach too makes sense for me. $\endgroup$
    – JWL
    Commented Sep 20, 2011 at 13:58
  • $\begingroup$ Check srivatsan his answer to see how this idea can be used in a full proof for this theorem. This is just the general idea for the proof. $\endgroup$
    – sxd
    Commented Sep 20, 2011 at 14:08
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Your proof isn't really right. What I mean is, for general matrices $A$ and $B$ the statement $$ A^n B^n = \prod A \prod B = \prod (AB) $$ may be wrong because $A$ and $B$ need not commute. I'd try doing this by induction instead.

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As Dimitri points out, your proof is incomplete. You can make it work in two ways:

  1. You can group the middle $AA^{-1}$ (remember that matrix product is associative). Noting that this equals $I$, the product simplifies to $A^{n-1} (A^{-1})^{n-1}$. You can then use induction to argue that $A^n (A^{-1})^{n}$ is $I$ for all $n$.

  2. This is slightly more general variant of the above trick. Suppose $A$ and $B$ are commuting matrices (i.e., $AB = BA$), then you can indeed use $$ A^n B^n = (AB)^n $$ guilt-free! (Notice that $A$ and $A^{-1}$ commute, so this justifies your proof now so you can justify the proof this way as well. Keep in mind that some justification is necessary, otherwise the proof is wrong or incomplete.) The proof of this fact also uses similar ideas; see if you can figure it out yourself.

In fact, if you have an arbitrary product of matrices consisting of $m$ $A$'s and $n$ $B$'s (and no other matrices), then you can show that this product equals $A^m B^n$. For example, if $A$ and $B$ commute, then $$ B^5ABA^2 B^{3} = A^{1+2} B^{5+1+3} = A^3 B^9. $$

A method by induction:

I leave you to verify that the result is true for the base case $n=0$. For the induction step, assume that $$ (A^{n-1})^{-1} = (A^{-1})^{n-1} .$$ We must now prove the claim for $n$. This follows from the chain of equalities: $$ \begin{eqnarray*} (A^n)^{-1} &=& (A \cdot A^{n-1})^{-1} \\ &\stackrel{({a})}{=}& (A^{n-1})^{-1} \cdot A^{-1} \\ &\stackrel{({b})}{=}& (A^{-1})^{n-1} \cdot A^{-1} \\ &=& (A^{-1})^{n}. \end{eqnarray*} $$ Be sure to justify each step, particularly the ones marked $(a)$ and $(b)$.

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    $\begingroup$ thanks. The Induction actually probably makes the proof more powerful and general. And for the "guilt-free!" $\endgroup$
    – JWL
    Commented Sep 20, 2011 at 14:22
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What I know is that if $A$ and $B$ are invertible then $AB$ is so. Then $A^n$ is invertible , I am agree with (mt_ ) by induction you can prove it: we know that

$n=1$ is correct $A^{-1}A=I$

if $n=k$ is correct ; $(A^{k})^{-1}A^k=I$ then we should show $(A^{k+1})^{-1}(A^{})^{k+1}=I$

we have $(A^{k+1})^{-1}A^{k+1}=(A^{k}A)^{-1}A^{k+1}=A^{-1}(A^k)^{-1}A^{k}A=A^{-1}(A)=I$

we know that $(AB)^{-1}=B^{-1}A^{-1}$

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