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

  • Thread starter trojansc82
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  • #1
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Homework Statement



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

(Ak)-1 = (A-1)k

Homework Equations



A-1A = I

The Attempt at a Solution



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

Answers and Replies

  • #2
1,086
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Your "relevant equation" is wrong. Perhaps you meant AA-1 = A-1A = I.
 
  • #3
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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.
 
  • #4
1,086
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As far as the solution, what is the inverse of (Ak)-1? What happens if you multiply the right-hand side by that inverse, as well?
 
  • #5
vela
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If you substitute Ak into the LHS of that equation, you get (Ak)-1(Ak). Can you show that equals I?
 
  • #6
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As far as the solution, what is the inverse of (Ak)-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 Ak the result will be I.

However, I'm having a hard time showing Ak(A-1)k = I.
 
  • #7
vela
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Just write it out:

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

And use the fact that matrix multiplication is associative.
 
  • #8
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This proof should almost certainly be done using induction.
 
  • #9
2,967
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This proof should almost certainly be done using induction.
by using the rule:

[tex]
(A B)^{-1} = B^{-1} A^{-1}
[/tex]

and the recursive relation:

[tex]
A^{k + 1} = A^{k} A
[/tex]
 
  • #10
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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)
 
  • #11
1,086
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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?
 
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  • #12
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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
 
  • #13
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Alright, so you have

[tex]\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} = ... [/tex]

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 Ak.
 
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  • #14
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Alright, so you have

[tex]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} = ... [/tex]

Does this help?
Yes, a little bit.

Does A-1 A-1 = A ?

If so, that would help with solving the proof.
 
  • #15
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[tex]A^{-1}A^{-1} \neq A[/tex]

However, I've edited my previous post with underbraces. Can you now see what I meant?
 
  • #16
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[tex]A^{-1}A^{-1} \neq A[/tex]

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 Ak, which would give me A-1 = A-1?
 
  • #17
1,086
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No no, you multiply both sides with Ak, not the inverse of Ak. Ak 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?
 
  • #18
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No no, you multiply both sides with Ak, not the inverse of Ak. Ak 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 Ak because (AAA...AA)(A-1A-1A-1...A-1A-1) will reduce to the identity matrix.
 
  • #19
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You are right in saying both side will equal the identity matrix, but not Ak. Remember, you are multiplying by Ak to actually get the identity matrix. So as you can see, on the right-hand side you did the following: (A-1)kAk = I. What does this tell you about (A-1)k in relation to Ak?
 
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  • #20
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You are right in saying both side will equal the identity matrix, but not Ak. Remember, you are multiplying by Ak to actually get the identity matrix. So as you can see, on the right-hand side you did the following: (A-1)kAk = I. What does this tell you about (A-1)k in relation to Ak?
It tells me that Ak is the inverse of (A-1)k, but I am not able to see it.
 
  • #21
HallsofIvy
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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, [itex](A^{-1})^1= A^{-1}= (A^1)^{-1}[/itex]

Now, assume that, for some k, [itex](A^{-1})^k= (A^k)^{-1}[/itex]

[itex] (A^{-1})^{k+1}= (A^{-1})^k A^{-1}= (A^k)^{-1}A^{-1}[/itex]
while [itex](A^k)^{-1}= (A^kA)^{-1}= A^{-1}(A^k)^{-1}[/tex]
 
  • #22
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It tells me that Ak 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 Ak. But we already know (Ak)-1 is an inverse of Ak. So can a matrix have two different inverses?

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

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