Proving Q=RP for Matrix A Using Commutative Property | Matrix Theory Homework

[arpitm08]

Homework Statement

If B≈A with (P^-1)*A*P = B and also (Q^-1)*A*Q = B, show that Q=RP where R is a nonsingular matrix that commutes with A.

Homework Equations

AR = RA

The Attempt at a Solution

Our professor told us that it is easier to just play around with these equations and get the answer. That breaking them down to their elements would be a ton of work. I tried multiplying by a bunch of various ways. First I changed the equivalence formulas to, AP=PB and AQ=QB, and I tried multiplying by R, because of the commuting factor, but I couldn't get anywhere doing that. If someone could give me a hint as to if I'm missing a property or something that could help me out, that'd be great. Thanks.

 

[tiny-tim]

hi arpitm08!

hint: what happens if you multiply either equation by Q ?

 

[Deveno]

i have an alternate approach:

note that B = P^-1AP is the same as: A = PBP^-1.

also, note that if R^-1AR = A, then AR = RA.

what happens if you evaluate (QP^-1)^-1A(QP^-1)?

 

[tiny-tim]

isn't that the same as mine, except using P instead of Q ?

 

[Deveno]

isn't that the same as mine, except using P instead of Q ?

your previous hint was:

"multiply either equation by Q".

i count 5 "=" in the OP's post, so it is unclear to me which two of them you mean. i suppose you mean:

1) P^-1AP = B
2) Q^-1AQ = B

note that "multiply by Q" is not unambiguously defined, since Mat(n,F) is a non-commutative monoid. presumably you meant "left-multiply by Q".

i think your IDEA is the same idea as mine, i think our hints are not.