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Math Help - An interesting problem on C=A+B

  1. #1
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    Question An interesting problem on C=A+B

    if A and B are symmetrical real matrices.
    C=A+B

    rc, ra, and rb are eigen values of C, A, and B.
    I found that for the first several eigen values, they could be approximated as rc~=ra+rb. Is there any reason?

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  2. #2
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    Quote Originally Posted by chenuo View Post
    if A and B are symmetrical real matrices.
    C=A+B

    rc, ra, and rb are eigen values of C, A, and B.
    I found that for the first several eigen values, they could be approximated as rc~=ra+rb. Is there any reason?

    Expect your answers!
    [b]If[/b ] A and B have the same eigenvectors, then the eigenvectors of C are the sums of corresponding egenvectors of A and B. "Corresponding" here meaning "for the same eigenvector". If Av= \lambda_A v and Bv= \lambda_B v, then (A+B)v= Av+ Bv= \lambda_A v+ \lambda_B v= (\lambda_A+ \lambda_B)v[/tex].

    Otherwise, I don't know how you are going to choose which eigenvalues of A and B to add to get eigenvalues of B.
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  3. #3
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    RE: HallsofIvy

    There's no any relationship between A and B and we can not assume that A and B have the same eigenvectors. The eigenvalues we chose are the largest ones. The equation rc=ra+rb does not hold excactly! It's just approximation.
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