Here's the question you clicked on:
PhoenixFire
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
Base 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.
write |dw:1349805212895:dw|
|dw:1349805366358:dw|
I 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: \((A_1A_2...A_k A_{k+1})^{-1}=A_{k+1}^{-1}A_k^{-1}..A_2^{-1}A_1^{-1}\)
For the final step, you can use the hints given to you by @experimentX
Please let me know if you require any further explanation.
I will attempt this in a few hours again... I have to go into Uni right now.
Okay 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?
yeah you are correct ... you can use this law to show for n=3,4,... up to n
for 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.
Excellent. I understand it now. Out of curiosity, is my direct proof also valid?
from this particular point, it's correct ... for n=2 2) If I choose my base step as n=2 and want to prove ------------
now show it for n=3 and n=4
Well 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.
yes 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 \( (A_1 A_2 ...A_nA_{n+1})^{-1} = A_{n_+1}^{-1}A_n^{-1} ...A_2^{-1}A_1^{-1} \)
doing that will make your matrix simple product of two matrices. use the simple trick and assumption to show that property.
Great! Thank you for your help!