Can anyone help?
If $A$ is an invertible $n \times n$ matrix and $X,Y$ are $n \times n$ matrices such that $X = AY$ and $X = (A\times A)Y$, does it follow that $X = Y$ ?
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What do you mean by $A\times A$? – user1551 Jun 10 '13 at 16:09
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Welcome to Math.SE! Thanks for bringing your questions here, hope you find it helpful and contribute to the site yourself. – gt6989b Jun 10 '13 at 16:11
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is this the same as asking if $AY=Y$? – user1783444 Jun 10 '13 at 16:12
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1@user1551 It's A squared, sorry to make you confused. I'm new here, thanks. – user81767 Jun 10 '13 at 16:14
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@user1783444 Hmm I guess so. – user81767 Jun 10 '13 at 16:17
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I don't think you should start a question with "Can anyone help?", people will think "help with what? you haven't even said it yet!". – JMCF125 Jun 10 '13 at 17:11
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We have $X = AY = A^2 Y$, which implies $0 = A^2 Y - AY = (A-I) AY$.
Now since A is invertible, either $A=I$ or $Y=0$. In either case, $X=AY=Y$.
gt6989b
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You got that $(A-I)Y=0$, ie. $AY=Y$, but how do you deduce that $A=I$ or $Y=0$? – Seirios Jun 10 '13 at 16:21
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Well, thanks. But what if we have BX=AY and B^2 X = A^2 Y? Does it still follow the same? – user81767 Jun 10 '13 at 16:24
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$X=AY$
$\Rightarrow Y=A^{-1}X=A^{-1}A^2Y=AY=X$
Abhra Abir Kundu
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Yeah thanks :) Hmm but I don't know why I feel this method will not bring me to solve the more general statement :
If B^k X = A^k Y for k = 1,2,...,2n, then X=Y
– user81767 Jun 10 '13 at 16:36 -
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We have $X=A^2Y=AY.$ Now, $$A^2Y=AY\Rightarrow AY=Y$$ Since $A$ is invertible. Again $AY=X$, hence $X=Y.$
pritam
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