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Characteristic polynomial

  1. Apr 29, 2005 #1
    So, I have stared at this for a while:
    Notation: Q' - inverse of Q, != stands for "not equal";

    Suppose A and B are nxn matrices such that A = QBQ' for some invertible matrix Q. Prove that A and B have the same characteristic polynomials

    I can prove that they have the same determinant, but that is about it. I know that charact. polyn. looks like so:

    det(A - I*lambda) = det(B - I*lambda)
    det(QBQ' - I*lamda) = det(B - I*lambda)

    It would be equal if Q and Q' cancel out, but isn't it true that det(A + B) != det(A) + det(B).
    I am not sure where to go from here. Is it correct to multiply expressions inside parenthesis by something on both sides? even if it's in determinant :uhh:
    I am studying for the final, so any help is appreciated more than ever.
     
  2. jcsd
  3. Apr 29, 2005 #2

    AKG

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    det(QBQ' - kI) = det(QBQ' - Q(kI)Q'), where I'm using k instead of [itex]\lambda[/itex] because it's easier to type. Now, do you see why this equality holds?

    Q(kI)Q' = kQIQ' = k(QI)Q' = k(Q)Q' = kQQ' = k(I) = kI, so the equality does hold. Use the distributive property of matrix multiplaction twice, then use the fact that det(ABC) = det(A)det(B)det(C), and you should be done.
     
  4. Apr 29, 2005 #3
    Ohhh, OK. So you also used this identity: AI = IA.
    Thanks :smile:
     
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