Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
4  
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. –  sxd Sep 20 '11 at 13:19
    
It's correct if before writing the equality, you write $AA^{-1}=I=A^{-1}A$. –  Davide Giraudo Sep 20 '11 at 13:23
    
@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.) –  Srivatsan Sep 20 '11 at 14:10
2  
@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) –  Srivatsan Sep 20 '11 at 14:21
2  
(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!" –  Srivatsan Sep 20 '11 at 14:24

4 Answers 4

up vote 7 down vote accepted

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 $(b)$.

share|improve this answer
1  
thanks. The Induction actually probably makes the proof more powerful and general. And for the "guilt-free!" –  nemesisfixx Sep 20 '11 at 14:22

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!

share|improve this answer
    
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: $\prod_{i=1}^{n-1}A.(AA^{-1}).\prod_{i=1}^{n-1}=\prod_{i=1}^{n-1}A.(I).\prod_{i=‌​1}^{n-1}=...=A.(AIA^{-1}).A^{-1}=AIA^{-1}=AA^{-1}=I$. –  nemesisfixx Sep 20 '11 at 13:45
1  
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. –  sxd Sep 20 '11 at 13:54
    
Thanks, this clarifies the problem, and your approach too makes sense for me. –  nemesisfixx Sep 20 '11 at 13:58
    
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. –  sxd Sep 20 '11 at 14:08

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.

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.