# 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 $\mathbf{\Lambda_{s}} \in \mathbb{R}^{N-1 \times N-1}$ created through the infinite summation of matrices as:
$
\mathbf{\Lambda_{s}} = \sum_{n=1}^\infty \mathbf{\Lambda_{s(n)}}
$

where
$
\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
$
\alpha(n) &= \frac{|\Psi(n)|^2}{|Z(n\omega_s)|^2} \left(1 - \mathrm{Re}\{\zeta_n\} \right)\mathrm{Re}\{Z(n\omega_s)\}
$

$
\zeta_n = e^{-j2\pi\frac{n}{N}}
$

With the following definitions:
$
0 \leq d \leq 1
$

and
$
Z(n\omega_s) = jn\omega_s L + r
$

In general it is safe to approximate $|Z(n\omega_s)|^2 \approx n^2 \omega_s^2 L^2$ and therefore
$
\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 $L f_s r$ 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.

2. Originally Posted by FlyingCapacitor
Given the matrix $\mathbf{\Lambda_{s}} \in \mathbb{R}^{N-1 \times N-1}$ created through the infinite summation of matrices as:
$
\mathbf{\Lambda_{s}} = \sum_{n=1}^\infty \mathbf{\Lambda_{s(n)}}
$

where
$
\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
$
\alpha(n) &= \frac{|\Psi(n)|^2}{|Z(n\omega_s)|^2} \left(1 - \mathrm{Re}\{\zeta_n\} \right)\mathrm{Re}\{Z(n\omega_s)\}
$

$
\zeta_n = e^{-j2\pi\frac{n}{N}}
$

With the following definitions:
$
0 \leq d \leq 1
$

and
$
Z(n\omega_s) = jn\omega_s L + r
$

In general it is safe to approximate $|Z(n\omega_s)|^2 \approx n^2 \omega_s^2 L^2$ and therefore
$
\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 $L f_s r$ 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

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

(Here, I have used $n$ 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.

3. Dear Opalg,

Indeed. Thank you!