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Math Help - Diagonalization of Matrix A

  1. #1
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    Diagonalization of Matrix A

    Hi,

    A is a 2 by 2 matrix:

    a 1
    ab b

    i found the eigenvalues λ: 0 and a+b,
    the corresponding (column)eigenvectors:

    [1/√(1+a^2) -a/√(1+a^2)] & [1/b√(1+(1/b)^2) 1/√(1+(1/b)^2)]

    Wich gives matrix Q:

    1/√(1+a^2) 1/b√(1+(1/b)^2)

    -a/√(1+a^2) 1/√(1+(1/b)^2)

    Now i have to show that inverse(Q) * A * Q = Λ (diagonalization of A)

    I get inverse(Q): (found det(Q)= 1+a)

    1/(1+a)(1/√(1+a^2)) a/(1+a)√(1+a^2)

    -1/(1+a)(b√(1+(1/b)^2)) 1/(1+a)√(1+(1/b)^2)


    but i'm not getting the right anwser. What am i doing wrong? Is there a quicker way to do this?

    Please help me out.

    regards,

    Moljka.
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  2. #2
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    Re: Diagonalization of Matrix A

    here is what i would do:

    the eigenvector you have for a+b is a bit messy, in the denominator, √(1+(1/b)^2) = √((b/b)^2+(1/b)^2) = √[(1+b^2)/b^2],

    so you could have chosen (1/√(1+b^2), b/√(1+b^2)) as the eigenvector instead (which matches the first eigenvector in style, at least).

    now, re-calculate det(Q), i think your answer should have a's AND b's in it (when i did it, it was kinda ugly).

    you might also wish to rationalize your denominators, by multiplying the first eigenvector by √(1 + a^2)/√(1 + a^2), and similarly for b.
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  3. #3
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    Re: Diagonalization of Matrix A

    is there a way to prove inverse(Q) * A * Q = Λ for non symmetric matrices, like one can prove transpose(Q) * A *Q= Λ for symmetrix matrixes by premultiplying Aq1=λq1 by transpose(q) (q= a column vector)?
    This would also show that inverse(Q) * A * Q = Λ (diagonalization of A). Its impossible to premultiply Aq1=λq1 by inverse(q) since the inverse does not exist for a vector though.
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  4. #4
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    Re: Diagonalization of Matrix A

    for certain kinds of matrices, yes. but not every matrix is diagonalizable. for example,

    [1 1]
    [0 1]

    is not diagonalizable. this matrix has the eigenvalue 0, but any eigenvector is a multiple of (1,0), which only gives us one column

    for the change of basis matrix (and we need two).
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  5. #5
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    Re: Diagonalization of Matrix A

    Thank you for helping me out.

    after computing inverse(Q) * A * Q = Λ i got the matrix Λ:

    0 fcd(a+(a^2)b)+fce(1+ab)/(f^3)(d^3)(e^3)

    0 (-afd+abfb)(f^2)(d^2)(e)-(f^2)(e^2)(d)(fd+bfe)/(f^3)(d^3)(e^3)

    for convenience i substituted the following with the letters:
    c=√1+a^2
    e=b√1+(1/b)^2 ( or b√(b^2+1)/b^2)
    d=√1+(1/b)^2 ( or √(b^2+1)/b^2)
    f=1+a


    either that can some how be rewritten to:
    0 0

    0 a+b

    or once again i made an error somewere along the way.
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