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Tuesday, 16 August 2011
Math GRE - #15
Suppose $A$ and $B$ are invertible $n\times n$ matrices. If $A$ and $B$ are similar, which of the following statements are true?
$A-2I$ and $B-2I$ are similar.
$A$ and $B$ have the same trace.
$A^{-1}$ and $B^{-1}$ are similar matrices.
Choices:
1 only.
2 only.
3 only.
1 and 2 only.
1, 2, and 3 only.
Solution :
1, 2, and 3 are all true.
Recall that $A$ and $B$ are similar if $B=P^{-1}AP$ for some invertible matrix $P$.
Since $\text{tr}(AB)=\text{tr}(BA)$, we know that 2 is true due to: \[\text{tr}(B)=\text{tr}(P^{-1}AP)=\text{tr}(APP^{-1})=\text{tr}(A).\]
Moreover, we have \[B^{-1}=(P^{-1}AP)^{-1}=P^{-1}A^{-1}P.\] But by the definition of being similar, this means that $A^{-1}$ and $B^{-1}$ are similar. So 3 is true. Since only the last choice offers 2 and 3 as true, we know that the answer is the last choice: 1, 2, and 3 are all true.
Can you prove why 1 must be true?
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This one almost got me. For some reason I thought 1 wasn't true, but knew that 2 & 3 were, so I did the same thing you did - only the last one has both 2 & 3, so it must be right.
After thinking about it, we have
\[A-2I=PBP^{-1}-2PIP^{-1}=P(B-2I)P^{-1}\]
so $A-2I$ and $B-2I$ are similar by the same matrix $P$.
@8002959260108264927.0 Yup! That was exactly the way I thought about it too!
2 comments:
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This webpage is LaTeX enabled. To type in-line formulae, type your stuff between two '$'. To type centred formulae, type '\[' at the beginning of your formula and '\]' at the end.
After thinking about it, we have
\[A-2I=PBP^{-1}-2PIP^{-1}=P(B-2I)P^{-1}\]
so $A-2I$ and $B-2I$ are similar by the same matrix $P$.
Yup! That was exactly the way I thought about it too!
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