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Math Help - 2 questions about eigenvalues and eigenvectors

  1. #1
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    2 questions about eigenvalues and eigenvectors

    hi, i have a little problem...

    A. i need to prove that if v is an eigenvector which is in common for both matrices A and B, then v is also an eigenvector of AB and BA.

    B. i also need to prove that if A and B have an n number of eigenverctors in common which are linear independent from each other, then AB = BA.

    thanks in advance...
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  2. #2
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    Quote Originally Posted by vonflex1 View Post
    hi, i have a little problem...

    A. i need to prove that if v is an eigenvector which is in common for both matrices A and B, then v is also an eigenvector of AB and BA.

    B. i also need to prove that if A and B have an n number of eigenverctors in common which are linear independent from each other, then AB = BA.

    thanks in advance...
    For A)
    I will show v is an eigenvector of AB. For BA it is similar.
    "v is an eigenvector which is in common for both matrices A and B" \implies \exists \lambda, \mu : Av = \lambda v, Bv = \mu v \implies  (AB)v = A(Bv) = A(\mu v) = \mu(Av) = \mu(\lambda v) = (\mu \lambda) v

    For B)
    Do you know Diagonalisability?
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  3. #3
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    For B)
    Do you know Diagonalisability?

    yes
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  4. #4
    ynj
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    Let X1,X2...Xn be the n common eigenvectors, P=(X1 X2 .. Xn). P is reversable since X1..Xn are linear independent. Then AP=A(X1 ... Xn)=(AX1...AXn)=(r1X1 r2X2...rnXn)=PD1,where D1 is a diagonal matrix,r1..rn is eigenvalue. So A=PD1P^(-1).Similarily, B=PD2P^(-1).So AB=PD1D2P^(-1)=PD2D1P^(-1)=BA,since D1D2=D2D1 is very obvious.
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