1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?
     
  2. jcsd
  3. Dec 11, 2012 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Similar matrices
  1. Similar matrices (Replies: 4)

  2. Similar matrices (Replies: 3)

  3. Similar matrices (Replies: 1)

  4. Similar matrices (Replies: 1)

  5. Similar matrices (Replies: 1)

Loading...