Can a matrix have two different inverses?

[trojansc82]

Homework Statement

I'm attempting to prove a property of inverse matrices. The property is as follows:

(A^k)^-1 = (A^-1)^k

Homework Equations

A^-1A = I

The Attempt at a Solution

(A^-1)^k = A^-1 A^-1...A^-1 = (A^k)^-1
I postulated that you could factor a k based on A^-1 being multiplied k number of times.

 

[Ryker]

Your "relevant equation" is wrong. Perhaps you meant AA^-1 = A^-1A = I.

 

[trojansc82]

Your "relevant equation" is wrong. Perhaps you meant AA^-1 = A^-1A = I.

Yes I did. Thanks for the information, I'll edit my original post.

 

[Ryker]

As far as the solution, what is the inverse of (A^k)^-1? What happens if you multiply the right-hand side by that inverse, as well?

 

[vela]

If you substitute A^k into the LHS of that equation, you get (A^k)^-1(A^k). Can you show that equals I?

 

[trojansc82]

As far as the solution, what is the inverse of (A^k)^-1? What happens if you multiply the right-hand side by that inverse, as well?

I'm assuming that if I multiply both sides by A^k the result will be I.

However, I'm having a hard time showing A^k(A^-1)^k = I.

 

[vela]

Just write it out:

(AAA...AA)(A^-1A^-1...A^-1A^-1A^-1)

And use the fact that matrix multiplication is associative.

 

[Mark44]

This proof should almost certainly be done using induction.

 

[Dickfore]

This proof should almost certainly be done using induction.

by using the rule:

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

and the recursive relation:

$$A^{k + 1} = A^{k} A$$

 

[trojansc82]

Just write it out:

(AAA...AA)(A^-1A^-1...A^-1A^-1A^-1)

And use the fact that matrix multiplication is associative.

Here is what I have:



(AAA...AA)(A^-1A^-1...A^-1A^-1A^-1) =

(A A^-1 A A^-1 A A^-1... A A^-1 ) =

(A^-1A^-1...A^-1A^-1A^-1) (AAA...AA)

 

[Ryker]

You are doing commutativity, which in general does not hold for matrix multiplication, whereas vela meant associativity. What do you get if you group the inner A with the inner A^-1?

 

[trojansc82]

You are doing commutativity, which in general does not hold for matrix multiplication, whereas vela meant associativity. What do you get if you group the inner A with the inner A^-1?

You will get the identity matrix. I see how you can do that for the left side of the proof, but for the right side I don't see how you can get to the identity matrix with A^-1A

 

[Ryker]

Alright, so you have

$$\underbrace{AAA ... AA(A}_\textrm{k times} \underbrace{A^{-1})A^{-1}A^{-1}...A^{-1}A^{-1}A^{-1}}_\textrm{k times} = \underbrace{AAA ... AA}_\textrm{k - 1 times}(I) \underbrace{A^{-1}A^{-1}...A^{-1}A^{-1}A^{-1}}_\textrm{k - 1 times} = ...$$

Does this help? And don't worry about the left-hand side, you have proven that it is equal to the identity matrix when you multiplied it with A^k.

 

[trojansc82]

Alright, so you have

$$AAA ... AA(AA^{-1})A^{-1}A^{-1}...A^{-1}A^{-1}A^{-1} = AAA ... AA(I)A^{-1}A^{-1}...A^{-1}A^{-1}A^{-1} = ...$$

Does this help?

Yes, a little bit.

Does A^-1 A^-1 = A ?

If so, that would help with solving the proof.

 

[Ryker]

$$A^{-1}A^{-1} \neq A$$

However, I've edited my previous post with underbraces. Can you now see what I meant?

 

[trojansc82]

$$A^{-1}A^{-1} \neq A$$

However, I've edited my previous post with underbraces. Can you now see what I meant?

Yeah. I'm trying to follow the steps that were taken.

So I multiply both sides by the inverse of A^k, which would give me A^-1 = A^-1?

 

[Ryker]

No no, you multiply both sides with A^k, not the inverse of A^k. A^k is namely the inverse of left-hand side. But as you can see, when you group (AA^-1) = I, you get one less A and one less A^-1. So what is AIA^-1? And what happens if you continue grouping the inner matrices together like that?

 

[trojansc82]

No no, you multiply both sides with A^k, not the inverse of A^k. A^k is namely the inverse of left-hand side. But as you can see, when you group (AA^-1) = I, you get one less A and one less A^-1. So what is AIA^-1? And what happens if you continue grouping the inner matrices together like that?

Both sides should equal A^k because (AAA...AA)(A^-1A^-1A^-1...A^-1A^-1) will reduce to the identity matrix.

 

[Ryker]

You are right in saying both side will equal the identity matrix, but not A^k. Remember, you are multiplying by A^k to actually get the identity matrix. So as you can see, on the right-hand side you did the following: (A^-1)^kA^k = I. What does this tell you about (A^-1)^k in relation to A^k?

 

[trojansc82]

You are right in saying both side will equal the identity matrix, but not A^k. Remember, you are multiplying by A^k to actually get the identity matrix. So as you can see, on the right-hand side you did the following: (A^-1)^kA^k = I. What does this tell you about (A^-1)^k in relation to A^k?

It tells me that A^k is the inverse of (A^-1)^k, but I am not able to see it.

 

[HallsofIvy]

Did you notice Mark44's statement that this should be done by induction?
That way you can avoid writing those long lists of "A"s!

When n= 1, $(A^{-1})^1= A^{-1}= (A^1)^{-1}$

Now, assume that, for some k, $(A^{-1})^k= (A^k)^{-1}$

$(A^{-1})^{k+1}= (A^{-1})^k A^{-1}= (A^k)^{-1}A^{-1}$
while [itex](A^k)^{-1}= (A^kA)^{-1}= A^{-1}(A^k)^{-1}[/tex]

 

[Ryker]

It tells me that A^k is the inverse of (A^-1)^k, but I am not able to see it.

Exactly! Or more importantly, (A^-1)^k is the inverse of A^k. But we already know (A^k)^-1 is an inverse of A^k. So can a matrix have two different inverses?

And when you said you don't see it, what do you mean?