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Thread: [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 $\displaystyle \mathsf{T}$ be a linear operator on a finite-dimensional inner product space $\displaystyle \mathsf{V}$. Then the following statements are equivalent:

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

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

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

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

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

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

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

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

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

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

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

    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 $\displaystyle X$ be upper triangular and unitary, partition $\displaystyle X$ so that:

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

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

    Now consider:

    $\displaystyle X^HX$

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

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


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

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

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

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

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

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

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

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

    First of all, is that correct so far?

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

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

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

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

    continuing this way we see that for any $\displaystyle 1 \leq i \leq n,$ the only non-zero entry in the $\displaystyle i$-th row of $\displaystyle A$ is $\displaystyle a_{ii},$ and hence $\displaystyle A$ 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 $\displaystyle A$ and $\displaystyle C$ are unitary? Obviously they are upper-triangular, but how do I prove that if I arbitrarily partition $\displaystyle X$, we have $\displaystyle A,C$ unitary?

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

    and:

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

    so:

    $\displaystyle A^HA=I_{sz(A)}$

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

    It might be easier to see what is going on if you let $\displaystyle C$ be a $\displaystyle 1 \times 1$ 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; Aug 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 $\displaystyle A$.

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

    Inverse of unitary $\displaystyle A$ is $\displaystyle A^H$. But $\displaystyle A^H$ will be lower triangular. Since 'the inverse of an upper triangular matrix is upper triangular', $\displaystyle A^H$ must be upper triangular too. Thus $\displaystyle A^H$ is both upper triangular and lower triangular, thereby diagonal. So is $\displaystyle A$
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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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  8. #8
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    oh!! this post was really helpful. Thank you guys
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