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3x3 Matrix Proof Help

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SimonM
I'm fairly sure that powers of some matrix are commutative


I'm 70% sure it doesn't generally though. :s-smilie:

Oh and unless I'm missing something here couldn't you just do the first one by multiplying pre- and post- on both sides by the inverse of M?
ukdragon37
Don't think so. A=M2B=M3A=M^2 B=M^3 the matrices are different so I think it just falls into the rule above as post- and pre- multiplying would yield different answers. Caution is the better part of prudence (something like that :shifty: )

Matrix multiplication is associative, so if we use your examples for A and B, we get:
M2∗M3=(M∗M)∗(M∗M∗M)=(MM)MMM=MMMMM=MMM(MM)=M3∗M2M^2 * M^3 = (M*M) * (M*M*M) = (MM)MMM = MMMMM = MMM(MM) = M^3 * M^2
Lo and behold, AB=BAAB = BA.
Brook Taylor
Matrix multiplication is associative, so if we use your examples for A and B, we get:
M2∗M3=(M∗M)∗(M∗M∗M)=(MM)MMM=MMMMM=MMM(MM)=M3∗M2M^2 * M^3 = (M*M) * (M*M*M) = (MM)MMM = MMMMM = MMM(MM) = M^3 * M^2


Oh well that's me told :p:
Reply 23
Avatar for SimonM
SimonM
18
ukdragon37
I'm 70% sure it doesn't generally though. :s-smilie:

Oh and unless I'm missing something here couldn't you just do the first one by multiplying pre- and post- on both sides by the inverse of M?


Generally AB≠BA\mathbf{A B} \not= \mathbf{B A}

But AnAm=Am+n=AkAl\mathbf{A}^n \mathbf{A}^m = \mathbf{A}^{m+n} = \mathbf{A}^k \mathbf{A}^l (assuming m+n=k+l and m,n,l,k are positive integers)
Reply 24
Avatar for Guaglio
Guaglio
OP
what's this pre- and post- multiplying business?
Guaglio
what's this pre- and post- multiplying business?


Let A=B
Pre-multiplying by M means multiply by M in front on both sides.
MA=MB
Post-multiplying by M means multiply by M at the end on both sides.
AM=BM

The order matters since matrices are not generally commutative, i.e. AB =/= BA.
Reply 26
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SimonM
18
Guaglio
what's this pre- and post- multiplying business?


OK.

Since

AB≠BA\mathbf{A B} \not = \mathbf{B A}

It matters which side of an matrix which you multiply on. Therefore we say whether we are "pre-" or "post-" multiplying based on which side we are multiplying on
Reply 27
Avatar for Guaglio
Guaglio
OP
AHA!! I didn't know that was allowed...
Reply 28
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Guaglio
OP
btw the solution was pre- and post-multiplying both sides of the initial property given in the Q by M−1 M^{-1} . If you were interested. I don't know if that was what you'd done.

Thanks for the help anyway
Guaglio
btw the solution was pre- and post-multiplying both sides of the initial property given in the Q by M−1 M^{-1} . If you were interested. I don't know if that was what you'd done.

Thanks for the help anyway


ukdragon37
Oh and unless I'm missing something here couldn't you just do the first one by multiplying pre- and post- on both sides by the inverse of M?


At least I know I can do something right :p:
Reply 30
Avatar for Guaglio
Guaglio
OP
HAHAHA sorry I completely missed that! Thanks mate!

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