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Math Help - Linear Algebra

  1. #1
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    Linear Algebra

    Find a unitary matrix U so that for
    A = [ 2, 3-3i] [3+3i, 5],A = UDU^H, where D is diagonal

    Explain why the determinant of A has to have real determinant
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  2. #2
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    Re: Linear Algebra

    note that A is hermetian, so A = A*.

    note as well that A* = (UDU*)* = U**D*U* = UD*U*.

    this means that D* = U*A*U = U*AU = D.

    now det(A) = det(UDU*) = det(D). since D = diag(d11,d22),

    D* = diag(d11*,d22*), and det(D) = det(D*) thus means:

    d11d22 = (d11)*(d22)* = (d11d22)*

    hence det(A) = det(D) is real.

    in this particular case, of course, one can just compute det(A):

    det(A) = (2)(5) - (3+3i)(3-3i) = 10 + 9 - 9i2 = 10 + 9 - (-9) = 10 + 18 = 28

    *****

    the above proof can be adapted easily to any nxn Hermitian matrix.

    so how do we find such a matrix U?

    first, lets find the eigenvalues for A:

    det(xI - A) =

    \begin{vmatrix}x-2&-3+3i\\-3-3i&5-x \end{vmatrix}

    = (x - 2)(x - 5) - (-3+3i)(-3-3i) = x^2 - 7x + 10 - 18

    = x^2 - 7x - 8 = (x - 8)(x + 1)

    so the eigenvalues of A are -1 and 8. solving (A + I)u = 0 we get:

    u = (-1+i,1) as an eigenvector. normalizing u we get the unit vector ((-1+i)/√3,1/√3).

    solving (A - 8I)v = 0, we get:

    v = (1-i,2), which normalizes to the unit vector ((1-i)/√6,2/√6). note that <u,v> = 0, so we may take:

    U = (u,v)^* = \begin{bmatrix}\frac{-1-i}{\sqrt{3}}&\frac{1}{\sqrt{3}}\\ \frac{1+i}{\sqrt{6}}&\frac{2}{\sqrt{6}} \end{bmatrix}

    i leave it in your capable hands to show that UAU* =

    \begin{bmatrix}-1&0\\0&8 \end{bmatrix}

    which is, of course, diagonal.
    Last edited by Deveno; July 24th 2012 at 05:42 PM.
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