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

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

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

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

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

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

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

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

    Now put together (i) and (iii): $\displaystyle \frac{1}{u_{ii}}=\overline{u_{ii}}\Longleftrightar row\,\left|u_{ii}\right|^2=1$ , 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):$\displaystyle U^{-1}$ is upper triangular, but $\displaystyle U^{-1}=U^{*}$ , and $\displaystyle U^{*}$ is LOWER triangular $\displaystyle \Longrightarrow\,U^{*}$ is both upper and lower triangular and then it is diagonal, and thus also $\displaystyle U$ is.
    Tonio
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