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How would i show this?

  1. Dec 17, 2003 #1
    Ok well i have two questions.

    1) If B = P^-1AP and let X be an eigenvector of A corresponding to the eigenvalue y. Show that y is an eigenvalue of B and find a corresponding eigenvector.

    This is what i did.
    AX=yI and since B = P^-1Ap -> A = PBP^-1
    so (PBP^-1)X=yI
    Now this is the part where i get lost. Am i on the right track?

    2) If A and B are nxn matrices, A is invertable, show that BA is similar to AB.

    So BA = P^-1ABP because BA is similar to AB. But im kinda lost now. I'm sure i have to do something with the fact that A is invertable. Umm... A^-1....
     
  2. jcsd
  3. Dec 17, 2003 #2

    Hurkyl

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    1)

    Here's your mistake; what you should have is Ax = yx.

    Recall that the definition is:

    [itex]\vec{v}[/itex] is an eigenvector of [itex]A[/itex] corresponding to eigenvalue [itex]\lambda[/itex] iff [itex]A\vec{v} = \lambda \vec{v}[/itex].


    2)

    Ok, you know to be similar, you need

    [tex]BA = P^{-1}ABP[/tex]

    What's the simplest guess as to what P should be to make this equation hold?
     
    Last edited: Dec 17, 2003
  4. Dec 17, 2003 #3
    lol

    i meant [itex]A\vec{v} = \lambda \vec{v}[/itex]. Stupid typo by me. But i still don't get it [b(]

    And for the other one, im lost as to what [itex]P[/itex] should be to make this equation hold. It says that [tex]A[/tex] and [itex]B[/itex] are nxn matrices. [itex]A[/itex] is invertable. show that [itex]BA[/itex] is similar to [itex]AB[/itex].. umm....
     
  5. Dec 17, 2003 #4

    Hurkyl

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    1)

    You know that you need to find something of the form [itex]B\vec{w}=\lambda\vec{w}[/itex]...

    but you have something of the form [itex]P^{-1}BP\vec{v} = y \vec{v}[/itex]...

    The first thing I notice is that the LHS must be the matrix [itex]B[/itex] times some vector... so can you rewrite what you do have in such a way that the LHS is [itex]B[/itex] times some vector?


    2)

    You need to find a [itex]P[/itex] such that:

    [tex]BA = P^{-1}ABP[/tex]


    The first thing I notice is that I need to have an [itex]A[/itex] on the right of [itex]B[/itex]... you have a [itex]B[/itex] on the right hand side, can you select a value for [itex]P[/itex] so that we have an [itex]A[/itex] on the right of the [itex]B[/itex]?
     
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