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

1. ## 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.

2. Originally Posted by FlyingCapacitor
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.

3. Dear Opalg,

Indeed. Thank you!