Linear Algebra- prove that A is similar to B then A inverse is similar to B invese.

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Homework Statement



If A and B are invertible matrices and B is similar to A, prove that B-1 is similar to A-1

Homework Equations





The Attempt at a Solution



Not sure how to do this.. I know that similar matrices have the same characteristic polynomials and the same eigenvalues and same determinant.. but I'm not sure how to tie that in with their inverses..
 

Answers and Replies

  • #2
Dick
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What's the definition of 'similar'?
 
  • #3
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There are some definitions and equations that you should have included amongst your relevant equations - namely, the definition of similarity and the formula for the inverse of a product of invertible matrices.
 
  • #4
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If A and B are similar there exists an invertible matrix P such that B = P-1AP.
 
  • #5
Fredrik
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Suppose X and Y are n×n matrices. Is there something that you can multiply XY with (either from the left or from the right) to get the identity matrix? When you have answered that, you have an explcity formula for (XY)-1, which you can use to rewrite the expression for B-1 that you already have.
 
  • #6
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How about the formula for the inverse of a product of matrices, where both matrices have inverses?
 
  • #7
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suppose x and y are n×n matrices. Is there something that you can multiply xy with (either from the left or from the right) to get the identity matrix? When you have answered that, you have an explcity formula for (xy)-1, which you can use to rewrite the expression for b-1 that you already have.

how about the formula for the inverse of a product of matrices, where both matrices have inverses?
(xy)(xy)-1 = (xy)-1(xy) = i?
 
  • #8
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No, what I'm asking about is: Do you know a formula for (AB)-1?

You should try to get into the habit of using caps for matrices. Using i for the identity matrix could easily be interpreted as the imaginary unit i.
 
  • #9
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( AB )-1 = ( B -1 A -1)?

My letters keep getting lower cased on their own??
 
  • #10
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So then I get

(AB)-1 = B-1A-1
(A(P-1AP))-1 = B-1A-1
(A(P-1AP))-1A = B-1A-1A
B-1 = (A(P-1AP))-1A

..?
 
  • #11
vela
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Start with the definition of similarity, [itex]B=P^{-1}AP[/itex], and invert both sides using the formula for the inverse of a product of matrices.
 
  • #12
Fredrik
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In other words, figure out how to generalize [itex](XY)^{-1}[/itex] to three matrices (What is [itex](XYZ)^{-1}[/itex]?), and use it to express [itex]B^{-1}[/itex] in a more useful way.
 
  • #13
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Oh this is kind of brilliant
B = P-1AP
(B)-1 = (P-1AP)-1
B-1 = P-1A-1(P-1)-1
B-1 = P-1A-1P

Oh man how are you guys so smart
 
  • #14
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For this next question it asks me to prove that if B is similar to A, then BT is similar to AT.. so I try the same thing but I get

BT = PTAT(PT)-1?
 
  • #15
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Looks right to me. If you want it in the form BT=Q-1AQ, then take Q=(PT)-1
 
  • #16
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Oh ok great that makes sense
 

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