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Unitarily equivalent

  1. Mar 29, 2012 #1
    I've had the flu all week.

    Of course, the book defines unitary equivalent, but it doesn't talk about an efficient method of determining if two matrices are unitarily equivalent.

    Is there an efficient way to determine if these matrices are unitarily equivalent?

    [itex]
    \begin{bmatrix}
    0 & 1 & 0\\
    -1 & 0 &0 \\
    0 &0 &1
    \end{bmatrix}[/itex]

    [itex]
    \begin{bmatrix}
    1 & 0 & 0\\
    0 & i &0 \\
    0 &0 &-i
    \end{bmatrix}[/itex]
     
  2. jcsd
  3. Mar 29, 2012 #2

    micromass

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    You can easily find the eigenvalues, no?
     
  4. Mar 29, 2012 #3
    How does that relate to unitary equivalence?
     
  5. Mar 29, 2012 #4
    I found that A and B are unitarily equivalent if they have the same sets of eigenvalues, counting multiplicity.

    A = P*BP (unitarily equivalent)

    det(A) = det(P*BP) = det(P*)det(B)det(P) = det(P*)det(P)det(B) = det(B)
    det(A) = det(B)

    Their characteristic polynomials must be equal.
     
  6. Mar 29, 2012 #5

    HallsofIvy

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    No, that is not true. The matrices
    [tex]A= \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}[/tex]
    and
    [tex]B= \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}[/tex]
    have the same eigenvalues (1 with multiplicity two) but are not unitarily equivalent because they do not have the same eigenvectors. A has every vector as eigenvector while B has only multiples of <1, 0> as eigenvectors.

    Two matrices are "unitarily equivalent" if and only if they have the same eigenvalues and the same corresponding eigenvectors.

     
  7. Mar 29, 2012 #6
    Okay, so I found the eigenvalues of each of the matrices: 1, -i, +i. Now I have the tedious job of finding the eigenvectors. -_-
     
  8. Mar 29, 2012 #7

    Dick

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    Halls is definitely wrong to say that they have to have the same eigenvectors. You just have to have the same number of linearly independent eigenvectors for every eigenvalue. You have three distinct eigenvalues. That means you don't have to compute the eigenvectors. Why?
     
    Last edited: Mar 29, 2012
  9. Mar 29, 2012 #8
    Ah, you're right. The dimensions of the eigenspaces are equal - 3.
     
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