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 2 years ago
Prove by induction that \[(A_1A_2...A_n)^{1}={A_n}^{1}...{A_2}^{1}{A_1}^{1} \]\[\forall n\in \mathbb{N}\] and A_i for i=1, 2, ... are invertible matrices of same dimensions
 2 years ago
Prove by induction that \[(A_1A_2...A_n)^{1}={A_n}^{1}...{A_2}^{1}{A_1}^{1} \]\[\forall n\in \mathbb{N}\] and A_i for i=1, 2, ... are invertible matrices of same dimensions

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PhoenixFire
 2 years ago
Best ResponseYou've already chosen the best response.0Base step: Let n=1 \[(A_1)^{1}=A_1^{1}\]\[A_1^{1}=A_1^{1}\] Inductive Hyp. : Assume the statement holds true for n=m for some m in Natural Numbers. Inductive Step: Let n=m+1\[(A_1A_2...A_mA_{m+1})^{1}=A_{m+1}^{1}A_m^{1}...A_2^{1}A_1^{1}\]\[(A_1A_2...A_mA_{m+1})^{1}=A_{m+1}^{1}(A_1A_2...A_m)^{1}\] That's about as far as I got. Something doesn't seem right, and the Inductive Step isn't making sense as I don't know how to continue.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1write dw:1349805212895:dw

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1349805366358:dw

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1I would first show that: \((A_1A_2)^{1}=A_2^{1}A_1^{1}\) Then I would assume it is true for n=k., i.e.: \((A_1A_2...A_k)^{1}=A_k^{1}..A_2^{1}A_1^{1}\) And finally I would show that this also leads to: \((A_1A_2...A_k A_{k+1})^{1}=A_{k+1}^{1}A_k^{1}..A_2^{1}A_1^{1}\)

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1For the final step, you can use the hints given to you by @experimentX

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1Please let me know if you require any further explanation.

PhoenixFire
 2 years ago
Best ResponseYou've already chosen the best response.0I will attempt this in a few hours again... I have to go into Uni right now.

PhoenixFire
 2 years ago
Best ResponseYou've already chosen the best response.0Okay I have 2 questions now. 1) Could I not state that from my Inductive Hypothesis (the statement is true for n=k) and let \[B_1=(A_1A_2...A_k)\]\[B_2=A_{k+1}\] and then show: \[(A_1A_2..A_nA_{n+1})^{1}=B_2^{1}B_1^{1}\] From Ind. Hyp. with an n=2 \[(A_1A_2..A_nA_{n+1})^{1}=(B_1B_2)^{1}\]And then substitute in B1 and B2 again to result in an equality of Left hand side and Right hand side? 2) If I choose my base step as n=2 and want to prove \[(A_1A_2)^{1}=A_2^{1}A_1^{1}\]Could I use a direct proof: Multiply both sides by A1A2\[(A_1A_2)^{1}A_1A_2=A_2^{1}A_1^{1}A_1A_2\]\[I=A_2^{1}A_2\]\[I=I\]Proving it for n=2. Am I correct to do it in this way?

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1yeah you are correct ... you can use this law to show for n=3,4,... up to n

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1for the last step, assume it to be true upto n, show that it is true for n+1 then this must be true for any n inductively.

PhoenixFire
 2 years ago
Best ResponseYou've already chosen the best response.0Excellent. I understand it now. Out of curiosity, is my direct proof also valid?

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1from this particular point, it's correct ... for n=2 2) If I choose my base step as n=2 and want to prove 

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1now show it for n=3 and n=4

PhoenixFire
 2 years ago
Best ResponseYou've already chosen the best response.0Well for 3 and 4 I'd use pretty much the same direct proof to reduce it to an Identity on both side proving an equality.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1yes you can do that ... let it assume to be true for some n. then show it to be true for n+1 ..... you can do this by assuming \( A_1 A_2 ...A_n = A\) so that \( A_1 A_2 ...A_n A_{n+1} = A\times A_{n+1}\) Now we assume \( (A_1 A_2 ...A_n)^{1} = A_n^{1} ...A_2^{1}A_1^{1} \) we show that \( (A_1 A_2 ...A_nA_{n+1})^{1} = A_{n_+1}^{1}A_n^{1} ...A_2^{1}A_1^{1} \)

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1doing that will make your matrix simple product of two matrices. use the simple trick and assumption to show that property.

PhoenixFire
 2 years ago
Best ResponseYou've already chosen the best response.0Great! Thank you for your help!
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