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Math Help - Unitary and upper triangular?

  1. #1
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    Unitary and upper triangular?

    Question: prove that a matrix that is both unitary and upper triangular must be diagonal.

    Call the matrix A. By the fact that it is unitary, AA^* = A^*A = I_n. In other words, \sum_{i=1}^n a_{ji} a^*_{ji} = 1 for all j=1,...,n where a^* is the complex conjugate. Additionally, \sum_{i=1}^n a_{ji} a^*_{ki} = 0 for j \neq k.

    And I know that A* is lower triangular. Any hints, please? Thanks.
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    Quote Originally Posted by Last_Singularity View Post
    Question: prove that a matrix that is both unitary and upper triangular must be diagonal.

    Call the matrix A. By the fact that it is unitary, AA^* = A^*A = I_n. In other words, \sum_{i=1}^n a_{ji} a^*_{ji} = 1 for all j=1,...,n where a^* is the complex conjugate. Additionally, \sum_{i=1}^n a_{ji} a^*_{ki} = 0 for j \neq k.

    And I know that A* is lower triangular. Any hints, please? Thanks.

    You're almost there, though your notation perhaps makes things a little messier:

    U=(u{_ij}) is unitary, so:

    (i) \Longrightarrow\,U^{*}=U^{-1}

    (ii) \sum\limits_{k=i}^n(u_{ik}\overline{u_{ik}}=)\left  |u_{ik}\right|^2=1

    (iii) The inverse of an upper triangular matrix U=(u_{ij}) is also upper triangular with diagonal elements =u_{ii}^{-1}=\frac{1}{u_{ii}}\,,\,i=1,2,...,n

    Now put together (i) and (iii): \frac{1}{u_{ii}}=\overline{u_{ii}}\Longleftrightar  row\,\left|u_{ii}\right|^2=1 , and taking into account (ii) this gives us that all the elements outside of the main diagonal must be zero. Q.E.D.


    Or simpler, using your reasoning and (iii): U^{-1} is upper triangular, but U^{-1}=U^{*} , and U^{*} is LOWER triangular \Longrightarrow\,U^{*} is both upper and lower triangular and then it is diagonal, and thus also U is.
    Tonio
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