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Proof that eigenvalues are constants

  1. Apr 19, 2015 #1
    i suppose matrices of the form $$A_i=\left(\begin{array}[cc] ccos(x_i)&sin(x_i)\\sin(x_i)&-cos(x_i)\end{array}\right)$$

    And i consider the matrix

    $$C=A1\otimes A2\otimes 1_4-A1\otimes 1_4\otimes A2$$

    I would like to show that the eigenvalues of C are independent of the x
    i tried with mathematica but i have no proof.
     
  2. jcsd
  3. Apr 19, 2015 #2

    robphy

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    What is ##1_4##?
     
  4. Apr 19, 2015 #3
    The identity matrix of dimension 4 $$\left(\begin{array}[cccc]a1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{array}\right)$$
     
  5. Apr 19, 2015 #4
    What do you mean by ##\otimes##?
     
  6. Apr 20, 2015 #5
    The usual Kronecker product or tensor product
     
  7. Apr 20, 2015 #6
    First, your matrices ##A_i## are the reflection matrices: in particular that means that the spectrum of ##A_i## is ##\{-1, 1\}## and that ##A_i## are unitarily equivalent to $$A_0= \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right),$$ meaning that ##A=U_i A_0 U_i^{-1}##, where ##U_i## are unitary operators (with real entries). Matrices ##U_i## can be easily computed, but it does not matter here.

    Since the spectrum ##\sigma(A\otimes B)## is the product of ##\sigma(A)## and ##\sigma (B)## you just need to prove that the spectrum of ##A_2\otimes I_4 - I_4\otimes A_2## does not depend on ##x_2##. To see that you can notice that ##I_4 = I_2\otimes I_2##, so $$A_2\otimes I_4 - I_4\otimes A_2 = A_2\otimes I_2\otimes I_2 - I_2\otimes I_2\otimes A_2, $$ and the latter operator is unitarily equivalent to $$A_0\otimes I_2\otimes I_2 - I_2\otimes I_2\otimes A_0$$ (can you see why?)

    So the spectrum does not depend on ##x_2##.
     
  8. Apr 21, 2015 #7
    To show this I think we multiply left by $$U\otimes U\otimes U$$ and right by the inverses of U ?

    Thanks a lot.
     
  9. Apr 22, 2015 #8
    If by ##U## you mean ##U_2##, then yes, you are correct. You can also use ##U_2\otimes I_2\otimes U_2##.
     
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