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Homework Help: Similar matrices

  1. Dec 11, 2012 #1
    If A and B are similar matrices, show that A has an inverse IFF B is invertible.

    A=P^-1 * B * P
    Where P is an invertible matrix.

    (A)^-1 = (P^-1 * B * P)^-1
    A^-1 = P^-1 * B^-1 * P

    Does this show what the question wanted?
     
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  3. Dec 11, 2012 #2

    Dick

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    Not until you explain in words how it does.
     
  4. Dec 11, 2012 #3
    This shows that B^-1 is similar to A^-1?


    I forgot to put this in the statement, the problem specifically asked to "use the "definition" of similar matrices to show this."

    My original proof was that if A and b are similar then they must have the same determinant. If B is not invertible then it's determinant is 0 and A will have a determinant of 0 and is therefore not invertible. If B is intervible then it's determinant is not 0 and....A is invertible. But the teacher said I needed to use the definition.
     
  5. Dec 11, 2012 #4

    Dick

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    Ok, then use the definition. If A is invertible then A^(-1) exists. Use what you've got to give a formula for B^(-1). The show it works. And vice versa.
     
  6. Dec 11, 2012 #5
    Alright so, after this:
    (A)^-1 = (P^-1 * B * P)^-1
    A^-1 = P^-1 * B^-1 * P
    P^A^-1*P^-1=B^-1

    Ugh, I guess I don't know what I need to do or maybe that I already have all the stuff, but not have worded it.
     
  7. Dec 11, 2012 #6

    Dick

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    Ok, so you are claiming B^(-1)=PA^(-1)P^(-1). Check it. B=PAP^(-1) right, from your definition? Multiply your expression for B^(-1) by the expression for B using those forms. What do you get? How does that prove part of what you want?
     
    Last edited: Dec 11, 2012
  8. Dec 11, 2012 #7
    Wait from the definition, doesn't B=P^-1 * A *P?
     
  9. Dec 11, 2012 #8

    Dick

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    No. P is a fixed matrix once you've used the definition. You said A=P^-1 * B * P. So B=PAP^(-1). You are misunderstanding what the definition of similarity means.
     
  10. Dec 11, 2012 #9
    (A)^-1 = (P^-1 * B * P)^-1
    A^-1 = P^-1 * B^-1 * P
    P^A^-1*P^-1=B^-1


    B*P^A^-1*P^-1=B*B^-1=I
    PAP^(-1)*P^A^-1*P^-1=I
    PA*I*A^-1*P^-1=I
    PP^-1=I
    I=I


    Ok, so you are claiming B^(-1)=PA^(-1)P^(-1). Check it. B=PAP^(-1) right, from your definition? Multiply your expression for B^(-1) by the expression for B using those forms. What do you get? How does that prove part of what you want?
     
  11. Dec 11, 2012 #10

    Dick

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    Any words to got along with all of those typos to indicate whether you get it or not? This sequence is rightish. PAP^(-1)*P^A^-1*P^-1=PA*I*A^-1*P^-1=PP^-1=I. So?
     
  12. Dec 11, 2012 #11
    If P*P^-1=I Then that means P is invertible. I think I'm just stating the obvious. This showed that multipling B^-1 by B gives the identity matrix which means B is invertible.
     
  13. Dec 11, 2012 #12

    Dick

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    You've shown something you think is B^(-1) times B is I. That proves that the thing that you think is B^(-1) is, in fact, B^(-1). If CB=I then C is B^(-1). I think this is how the exercise wants you to show it.
     
  14. Dec 12, 2012 #13

    HallsofIvy

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    By the way, in a more abstract sense, a matrix always represents a linear transformation in some basis. A "similar" matrix represents the same linear transformation in a different basis. Of course, whether or not a linear transformation is invertible is independent of the specific basis in which it is represented.
     
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