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Thread: Unitary and triangular = diagonal matrix

  1. #1
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    Unitary and triangular = diagonal matrix

    can anybody help me with this question?

    Show that if a matrix A is both triangular and unitary, then it is diagonal.

    thanks
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  2. #2
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    Quote Originally Posted by Bucephalus View Post

    can anybody help me with this question?

    Show that if a matrix A is both triangular and unitary, then it is diagonal.

    thanks
    i'll assume that the matrix $\displaystyle A=[a_{ij}], \ 1 \leq i,j \leq n,$ is upper triangular. the lower triangular case is the same. proof is by induction over $\displaystyle n$: for $\displaystyle n=2$ it's easy. now if $\displaystyle [b_{ij}]=AA^*=I_n,$ where $\displaystyle A^*$

    is the complex conjugate of $\displaystyle A,$ then we'll have $\displaystyle b_{in}=0, \ \forall \ i \leq n-1,$ but $\displaystyle b_{in}=a_{in} \overline{a_{nn}}$ and $\displaystyle a_{nn} \neq 0.$ therefore $\displaystyle b_{in}=0, \ \forall \ i \leq n-1.$ now apply the induction hypothesis for the $\displaystyle (n-1) \times (n-1)$

    matrix $\displaystyle C=[a_{ij}], \ 1 \leq i,j \leq n-1$ to finish the proof.
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  3. #3
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    then we'll have $\displaystyle b_{in}=0, \ \forall \ i \leq n-1,$ but $\displaystyle b_{in}=a_{in} \overline{a_{nn}}$ and $\displaystyle a_{nn} \neq 0.$ therefore $\displaystyle b_{in}=0, \ \forall \ i \leq n-1.$


    I don't really understand how

    $\displaystyle b_{in}=a_{in} \overline{a_{nn}}$ and $\displaystyle a_{nn} \neq 0.$
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  4. #4
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    $\displaystyle b_{in}=a_{in} \overline{a_{nn}}$ and $\displaystyle a_{nn} \neq 0.$
    From what I'm reading here is that for a say a 2X2 matrix, the element
    $\displaystyle b_{i2}$ is the $\displaystyle a_{i2}$ element multiplied by the
    $\displaystyle b_{22}$ element.

    Is that correct?
    i'm going to try that on paper.
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  5. #5
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    SOrry, I mean the conjugate of the $\displaystyle a_{22}$ element.
    not $\displaystyle b_{aa}$ element.

    Don't worry,I will do some paper work here.
    I think I'm starting to understand some of your proof using a 2X2 Identity matrix.

    thanks for your response.
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