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Math Help - [SOLVED] Prove that a matrix that is both unitary and upper triangular must be diagon

  1. #1
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    [SOLVED] Prove that a matrix that is both unitary and upper triangular must be diagon

    Prove that a matrix that is both unitary and upper triangular must be a diagonal matrix.

    In case you don't remember what a unitary matrix is, please refer to wikipedia.

    Possibly relevant theorem from the textbook:

    Theorem 6.18. Let \mathsf{T} be a linear operator on a finite-dimensional inner product space \mathsf{V}. Then the following statements are equivalent:

    (a) \mathsf{T}\mathsf{T}^*=\mathsf{T}^*\mathsf{T}=\mat  hsf{I}.

    (b) \langle\mathsf{T}(x),\mathsf{T}(y)\rangle=\langle x,y\rangle for all x,y\in\mathsf{V}.

    (c) If \beta is an orthonormal basis for \mathsf{V}, then \mathsf{T}(\beta) is an orthonormal basis for \mathsf{V}.

    (d) There exists an orthonormal basis \beta for \mathsf{V} such that \mathsf{T}(\beta) is an orthonormal basis for \mathsf{V}.

    (e) \lVert\mathsf{T}(x)\rVert=\lVert x\rVert for all x\in\mathsf{V}.

    Every time I start into this, it turns into something horribly messy. I feel like there's a trick I'm missing. Any thoughts? Or perhaps a full solution?

    Thanks!
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  2. #2
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    Quote Originally Posted by hatsoff View Post
    Prove that a matrix that is both unitary and upper triangular must be a diagonal matrix.

    In case you don't remember what a unitary matrix is, please refer to wikipedia.

    Possibly relevant theorem from the textbook:

    Theorem 6.18. Let \mathsf{T} be a linear operator on a finite-dimensional inner product space \mathsf{V}. Then the following statements are equivalent:

    (a) \mathsf{T}\mathsf{T}^*=\mathsf{T}^*\mathsf{T}=\mat  hsf{I}.

    (b) \langle\mathsf{T}(x),\mathsf{T}(y)\rangle=\langle x,y\rangle for all x,y\in\mathsf{V}.

    (c) If \beta is an orthonormal basis for \mathsf{V}, then \mathsf{T}(\beta) is an orthonormal basis for \mathsf{V}.

    (d) There exists an orthonormal basis \beta for \mathsf{V} such that \mathsf{T}(\beta) is an orthonormal basis for \mathsf{V}.

    (e) \lVert\mathsf{T}(x)\rVert=\lVert x\rVert for all x\in\mathsf{V}.

    Every time I start into this, it turns into something horribly messy. I feel like there's a trick I'm missing. Any thoughts? Or perhaps a full solution?

    Thanks!
    Let X be upper triangular and unitary, partition X so that:

    X=\left[  <br />
\begin{array}{cc}A&B\\0&C\end{array}<br />
\right]

    Where A and C are upper triangular square unitary matrices (not necessarily the same size).

    Now consider:

    X^HX

    CB
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  3. #3
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    So let's make sure I understand the proof correctly...

    Choose X an upper-triangular, unitary matrix. Partition X so that


    X=\left[\begin{array}{cc}A&B\\0&C\end{array}\right],

    with A,C upper-triangular unitary matrices. Then we have

    XX^*=\left[\begin{array}{cc}A&B\\0&C\end{array}\right]\left[\begin{array}{cc}A^*&0\\B^*&C^*\end{array}\right]

    =\left(\left[\begin{array}{cc}A&0\\0&C\end{array}\right]+\left[\begin{array}{cc}0&B\\0&0\end{array}\right]\right)\left(\left[\begin{array}{cc}A^*&0\\0&C^*\end{array}\right]+\left[\begin{array}{cc}0&0\\B^*&0\end{array}\right]\right)

    =\left[\begin{array}{cc}A&0\\0&C\end{array}\right]\left[\begin{array}{cc}A^*&0\\0&C^*\end{array}\right]+\left[\begin{array}{cc}0&B\\0&0\end{array}\right]\left[\begin{array}{cc}A^*&0\\0&C^*\end{array}\right] +\left[\begin{array}{cc}A&0\\0&C\end{array}\right]\left[\begin{array}{cc}0&0\\B^*&0\end{array}\right]+\left[\begin{array}{cc}0&B\\0&0\end{array}\right]\left[\begin{array}{cc}0&0\\B^*&0\end{array}\right]

    =\left[\begin{array}{cc}AA^*&0\\0&CC^*\end{array}\right]+\left[\begin{array}{cc}0&BC^*\\0&0\end{array}\right]+\left[\begin{array}{cc}0&0\\CB^*&0\end{array}\right]+\left[\begin{array}{cc}BB^*&0\\0&0\end{array}\right]

    =\left[\begin{array}{cc}AA^*+BB^*&BC*\\CB^*&CC^*\end{arra  y}\right]=I_n.

    So BC^*=CB^*=0 \Rightarrow BC^*C=B=0.

    First of all, is that correct so far?

    If yes, then I have one remaining question: Why can we assume A and C are unitary? Obviously they are upper-triangular, but how do I prove that if I arbitrarily partition X, we have A,C unitary?

    Thanks!
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  4. #4
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    let A=(a_{ij}) be an n \times n unitary and upper-triangular matrix. since A is upper triangular, we have \det A = a_{11}a_{22} \cdots a_{nn} \neq 0. thus a_{ii} \neq 0, \ \forall i. now let e_j \in \mathbb{C}^n be the column with 1 in the j-th

    entry and 0 everywhere else. note that Ae_j is the j-th column of A. let 1 < j \leq n. then 0=<e_1,e_j>=<Ae_1,Ae_j>=a_{11}a_{1j}, and thus a_{1j}=0, for all 1 < j \leq n. so all the entries in the first

    row of A except a_{11} are 0. next, let 2 < j \leq n. then 0=<e_2,e_j>=<Ae_2, Ae_j>=a_{22}a_{2j}, and thus a_{2j}=0, for all 2 < j \leq n. so all the entries in the second row of A except a_{22} are 0.

    continuing this way we see that for any 1 \leq i \leq n, the only non-zero entry in the i-th row of A is a_{ii}, and hence A is diagonal.
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  5. #5
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    Quote Originally Posted by hatsoff View Post
    If yes, then I have one remaining question: Why can we assume A and C are unitary? Obviously they are upper-triangular, but how do I prove that if I arbitrarily partition X, we have A,C unitary?

    Thanks!
    X^H=X=\left[\begin{array}{cc}A^H&0\\B^H&C^H\end{array}\right]

    and:

    X^HX=\left[\begin{array}{cc}A^H&0\\B^H&C^H\end{array}\right]\left[\begin{array}{cc}A&B\\0&C\end{array}\right] =\left[\begin{array}{cc}A^HA&A^HB\\B^HA&B^HB+C^HC\end{arr  ay}\right]=I_{sz(X)}

    so:

    A^HA=I_{sz(A)}

    and to show CC^H=I_{sz(C)} we need only consider XX^H

    It might be easier to see what is going on if you let C be a 1 \times 1 matrix (and this is all that is needed for the obvious induction required to complete the proof by this method)

    CB
    Last edited by CaptainBlack; August 4th 2009 at 11:21 PM.
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  6. #6
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    Quote Originally Posted by hatsoff View Post
    Prove that a matrix that is both unitary and upper triangular must be a diagonal matrix.
    Alternative:

    Let the matrix be A.

    If you know that 'the inverse of an upper triangular matrix is upper triangular', the proof is simpler.

    Inverse of unitary A is A^H. But A^H will be lower triangular. Since 'the inverse of an upper triangular matrix is upper triangular', A^H must be upper triangular too. Thus A^H is both upper triangular and lower triangular, thereby diagonal. So is A
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  7. #7
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    Thanks a lot! This is all very helpful.

    You guys really came through for me. I hope I can return the favor soon.
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    oh!! this post was really helpful. Thank you guys
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