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Thread: Proof that Matrix (found through summation of matrices) is positive definite

  1. #1
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    Proof that Matrix (found through summation of matrices) is positive definite

    Given the matrix $\displaystyle \mathbf{\Lambda_{s}} \in \mathbb{R}^{N-1 \times N-1} $ created through the infinite summation of matrices as:
    $\displaystyle
    \mathbf{\Lambda_{s}} = \sum_{n=1}^\infty \mathbf{\Lambda_{s(n)}}
    $
    where
    $\displaystyle
    \mathbf{\Lambda_{s(n)}} = \alpha(n)\left[
    \begin{array}{ccccc}
    1 & \cos(2\pi n \frac{1}{N}) & \cos(2\pi n \frac{2}{N}) & \cdots & \cos(2\pi n \frac{N-2}{N})\\
    \cos(2\pi n \frac{N-1}{N}) & 1 & \cos(2\pi n \frac{1}{N}) & \cdots & \cos(2\pi n \frac{N-3}{N})\\
    \cos(2\pi n \frac{N-2}{N}) & \cos(2\pi n \frac{N-1}{N}) & 1 & \cdots & \cos(2\pi n \frac{N-4}{N})\\
    \vdots & \vdots & \vdots & \ddots & \vdots \\
    %x_3 & x_4 & x_5 & x_0 & x_1\\
    \cos(2\pi n \frac{2}{N}) & \cos(2\pi n \frac{3}{N}) & \cos(2\pi n \frac{4}{N}) & \cdots & 1
    \end{array}
    \right]
    $
    and
    $\displaystyle
    \alpha(n) &= \frac{|\Psi(n)|^2}{|Z(n\omega_s)|^2} \left(1 - \mathrm{Re}\{\zeta_n\} \right)\mathrm{Re}\{Z(n\omega_s)\}
    $
    $\displaystyle
    \zeta_n = e^{-j2\pi\frac{n}{N}}
    $

    With the following definitions:
    $\displaystyle
    0 \leq d \leq 1
    $
    and
    $\displaystyle
    Z(n\omega_s) = jn\omega_s L + r
    $

    In general it is safe to approximate $\displaystyle |Z(n\omega_s)|^2 \approx n^2 \omega_s^2 L^2 $ and therefore
    $\displaystyle
    \alpha(n) \approx \frac{\sin^2(2\pi d)r}{n^4 \pi^4 L^2 f_s^2} \left( 1 - \cos(2\pi \frac{n}{N})\right)
    $
    The symbols $\displaystyle L f_s r $ are electrical engineering symbols for inductance, frequency and resistance and generally it is true that $\displaystyle Lf_s > 1$.

    I have tried my best but can not prove that $\displaystyle \mathbf{\Lambda_s}$ is positive definite. Any ideas would be appreciated!

    Thank you.
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by FlyingCapacitor View Post
    Given the matrix $\displaystyle \mathbf{\Lambda_{s}} \in \mathbb{R}^{N-1 \times N-1} $ created through the infinite summation of matrices as:
    $\displaystyle
    \mathbf{\Lambda_{s}} = \sum_{n=1}^\infty \mathbf{\Lambda_{s(n)}}
    $
    where
    $\displaystyle
    \mathbf{\Lambda_{s(n)}} = \alpha(n)\left[
    \begin{array}{ccccc}
    1 & \cos(2\pi n \frac{1}{N}) & \cos(2\pi n \frac{2}{N}) & \cdots & \cos(2\pi n \frac{N-2}{N})\\
    \cos(2\pi n \frac{N-1}{N}) & 1 & \cos(2\pi n \frac{1}{N}) & \cdots & \cos(2\pi n \frac{N-3}{N})\\
    \cos(2\pi n \frac{N-2}{N}) & \cos(2\pi n \frac{N-1}{N}) & 1 & \cdots & \cos(2\pi n \frac{N-4}{N})\\
    \vdots & \vdots & \vdots & \ddots & \vdots \\
    %x_3 & x_4 & x_5 & x_0 & x_1\\
    \cos(2\pi n \frac{2}{N}) & \cos(2\pi n \frac{3}{N}) & \cos(2\pi n \frac{4}{N}) & \cdots & 1
    \end{array}
    \right]
    $
    and
    $\displaystyle
    \alpha(n) &= \frac{|\Psi(n)|^2}{|Z(n\omega_s)|^2} \left(1 - \mathrm{Re}\{\zeta_n\} \right)\mathrm{Re}\{Z(n\omega_s)\}
    $
    $\displaystyle
    \zeta_n = e^{-j2\pi\frac{n}{N}}
    $

    With the following definitions:
    $\displaystyle
    0 \leq d \leq 1
    $
    and
    $\displaystyle
    Z(n\omega_s) = jn\omega_s L + r
    $

    In general it is safe to approximate $\displaystyle |Z(n\omega_s)|^2 \approx n^2 \omega_s^2 L^2 $ and therefore
    $\displaystyle
    \alpha(n) \approx \frac{\sin^2(2\pi d)r}{n^4 \pi^4 L^2 f_s^2} \left( 1 - \cos(2\pi \frac{n}{N})\right)
    $
    The symbols $\displaystyle L f_s r $ are electrical engineering symbols for inductance, frequency and resistance and generally it is true that $\displaystyle Lf_s > 1$.

    I have tried my best but can not prove that $\displaystyle \mathbf{\Lambda_s}$ is positive definite. Any ideas would be appreciated!

    Thank you.
    I can show that each of the matrices $\displaystyle \mathbf{\Lambda_{s(n)}}$ is positive (by which I mean positive semi-definite). You would then need to show that the series $\displaystyle \mathbf{\Lambda_{s}} = \sum_{n=1}^\infty \mathbf{\Lambda_{s(n)}}$ converges, and it would follow that $\displaystyle \mathbf{\Lambda_{s}}$ is positive (because positivity is preserved by taking sums and limits).

    Since the numerical coefficient $\displaystyle \alpha(n)$ (in the definition of $\displaystyle \mathbf{\Lambda_{s(n)}}$) is clearly positive, you are trying to show that the $\displaystyle n\times n$ matrix $\displaystyle M_\theta$ is positive for a certain value of $\displaystyle \theta$, where

    $\displaystyle M_\theta = \begin{bmatrix}1 & \cos\theta & \cos(2\theta) & \cdots & \cos\bigl((n-1)\theta\bigr)\\
    \cos\theta & 1 & \cos\theta & \cdots & \cos\bigl((n-2)\theta\bigr)\\
    \cos(2\theta) & \cos\theta & 1 & \cdots & \cos\bigl((n-3)\theta\bigr)\\
    \vdots & \vdots & \vdots & \ddots & \vdots \\
    \cos\bigl((n-1)\theta\bigr) & \cos\bigl((n-2)\theta\bigr) & \cos\bigl((n-3)\theta\bigr) & \cdots & 1 \end{bmatrix}$.

    (Here, I have used $\displaystyle n$ in place of your $\displaystyle N-1$, and I have used the fact that $\displaystyle \cos(2\pi-\theta) = \cos\theta$ to rewrite the entries below the diagonal.)

    In fact, the matrix $\displaystyle M_\theta$ is positive for all values of $\displaystyle \theta$. The quickest way to see this is to use complex numbers and to write $\displaystyle \omega = e^{i\theta}$. Then the complex conjugate and the inverse of $\displaystyle \omega$ are equal: $\displaystyle \overline{\omega} = \omega^{-1}$. Also, $\displaystyle \frac12(\omega^k + \overline{\omega}^k) = \cos k\theta$. Define a $\displaystyle 1\times n$ matrix $\displaystyle A_\omega$ by $\displaystyle A_\omega = \begin{bmatrix}1&\omega&\omega^2&\cdots&\omega^{n-1}\end{bmatrix}$. Then $\displaystyle A_\omega^*A_\omega$ is an $\displaystyle n\times n$ matrix whose $\displaystyle (j,k)$-element is $\displaystyle \overline{\omega}^j\omega^k = \omega^{k-j}$ (for $\displaystyle 0\leqslant j\leqslant n-1$ and $\displaystyle 0\leqslant k\leqslant n-1$). The matrix $\displaystyle \frac12(A_\omega^*A_\omega + A_{\overline{\omega}}^*A_{\overline{\omega}})$ has $\displaystyle (j,k)$-element $\displaystyle \frac12(\omega^{k-j} + \omega^{j-k}) = \cos\bigl((k-j)\theta\bigr)$ which is equal to the $\displaystyle (j,k)$-element of $\displaystyle M_\theta$.

    Therefore $\displaystyle M_\theta = \frac12(A_\omega^*A_\omega + A_{\overline{\omega}}^*A_{\overline{\omega}})$. But any matrix of the form $\displaystyle A^*A$ is positive, so it follows that $\displaystyle M_\theta$ is positive.
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  3. #3
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    Dear Opalg,

    Indeed. Thank you!
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