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Math Help - 2 proofs for unitary matrix/transformation prob.

  1. #1
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    2 proofs for unitary matrix/transformation prob.

    first one is if A and B are unitary, then prove (or disprove) A + B is unitary.

    so we have AA* = I and BB* = I.

    AA* + BB* = 2I

    Can you do this: (A + B)* = (A* + B*)?
    If you can, then we can have (A + B)(A + B)* = AA* + AB* + BA* + BB* = I
    so we would only need to prove AB* + BA* = -I. How do you do that?

    2nd proof:We have a real orthogonal n x n matrix A. If lamda is a complex eigenvalue of A, then prove the complex conjugate of lamda is also an eigenvalue of A. In other words, the non real eigenvalues of A occur in conjugate pairs.
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  2. #2
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    Quote Originally Posted by hashshashin715 View Post
    first one is if A and B are unitary, then prove (or disprove) A + B is unitary.


    Both I\,,\,-I are unitary, so...


    so we have AA* = I and BB* = I.

    AA* + BB* = 2I

    Can you do this: (A + B)* = (A* + B*)?
    If you can, then we can have (A + B)(A + B)* = AA* + AB* + BA* + BB* = I
    so we would only need to prove AB* + BA* = -I. How do you do that?

    2nd proof:We have a real orthogonal n x n matrix A. If lamda is a complex eigenvalue of A, then prove the complex conjugate of lamda is also an eigenvalue of A. In other words, the non real eigenvalues of A occur in conjugate pairs.

    Prove that, in general, if \lambda is an eigenvalue of A , then \overline{\lambda} is an

    eigenvalue of A^* (we assume all through that we use an orthonormal basis)

    Tonio
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  3. #3
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    For your first proof, note that

    A=\begin{bmatrix}1 &0\\ 0&1\end{bmatrix} and

    B=\begin{bmatrix}0 &1\\ 1&0\end{bmatrix} are both unitary (check this!).

    But

    A+B=\begin{bmatrix}1 &1\\ 1&1\end{bmatrix} is not unitary (why not?).
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by hashshashin715 View Post
    2nd proof:We have a real orthogonal n x n matrix A. If lamda is a complex eigenvalue of A, then prove the complex conjugate of lamda is also an eigenvalue of A. In other words, the non real eigenvalues of A occur in conjugate pairs.
    The hypothesis A orthogonal is irrelevant, we only need to know A is a real matrix. If \lambda is a complex eigenvalue of A\in \mathbb{R}^{n\times n} then, there exists 0\neq z\in \mathbb{C}^{n\times 1} such that Az=\lambda z . Taking conjugate, A\bar{z}=\bar{\lambda}\bar{z} with \bar{z}\neq 0 which implies that \bar{\lambda} is an eigenvalue of A .


    Fernando Revilla
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