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

    With the following definitions:
    <br />
0 \leq d \leq 1<br />
    and
    <br />
Z(n\omega_s) = jn\omega_s L + r<br />

    In general it is safe to approximate |Z(n\omega_s)|^2 \approx n^2 \omega_s^2 L^2 and therefore
    <br />
\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)<br />
    The symbols  L f_s r are electrical engineering symbols for inductance, frequency and resistance and generally it is true that  Lf_s > 1.

    I have tried my best but can not prove that \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  \mathbf{\Lambda_{s}} \in \mathbb{R}^{N-1 \times  N-1} created through the infinite summation of matrices as:
    <br />
\mathbf{\Lambda_{s}} = \sum_{n=1}^\infty \mathbf{\Lambda_{s(n)}}<br />
    where
    <br />
 \mathbf{\Lambda_{s(n)}} = \alpha(n)\left[<br />
           \begin{array}{ccccc}<br />
            1 & \cos(2\pi n \frac{1}{N}) & \cos(2\pi n \frac{2}{N}) & \cdots & \cos(2\pi n \frac{N-2}{N})\\<br />
        \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})\\<br />
            \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})\\<br />
        \vdots & \vdots & \vdots & \ddots & \vdots \\             <br />
%x_3 & x_4 & x_5 & x_0 & x_1\\<br />
            \cos(2\pi n \frac{2}{N}) & \cos(2\pi n \frac{3}{N}) & \cos(2\pi n \frac{4}{N}) & \cdots & 1    <br />
           \end{array}<br />
           \right]<br />
    and
    <br />
\alpha(n) &=  \frac{|\Psi(n)|^2}{|Z(n\omega_s)|^2} \left(1 - \mathrm{Re}\{\zeta_n\} \right)\mathrm{Re}\{Z(n\omega_s)\}<br />
    <br />
\zeta_n = e^{-j2\pi\frac{n}{N}}<br />

    With the following definitions:
    <br />
0 \leq d \leq 1<br />
    and
    <br />
Z(n\omega_s) = jn\omega_s L + r<br />

    In general it is safe to approximate |Z(n\omega_s)|^2 \approx n^2 \omega_s^2 L^2 and therefore
    <br />
\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)<br />
    The symbols  L f_s r are electrical engineering symbols for inductance, frequency and resistance and generally it is true that  Lf_s > 1.

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

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

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

    M_\theta = \begin{bmatrix}1 & \cos\theta & \cos(2\theta) & \cdots & \cos\bigl((n-1)\theta\bigr)\\<br />
\cos\theta & 1 & \cos\theta & \cdots & \cos\bigl((n-2)\theta\bigr)\\<br />
\cos(2\theta) & \cos\theta & 1 & \cdots & \cos\bigl((n-3)\theta\bigr)\\<br />
\vdots & \vdots & \vdots & \ddots & \vdots \\             <br />
\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 n in place of your N-1, and I have used the fact that \cos(2\pi-\theta) = \cos\theta to rewrite the entries below the diagonal.)

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

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

    Indeed. Thank you!
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