Aufgaben:Exercise 4.16: Eigenvalues and Eigenvectors

• Return to book

Three correlation matrices

Although the description of Gaussian random variables using vectors and matrices is actually only necessary and makes sense for more than  $N = 2$  dimensions,  here we restrict ourselves to the special case of two-dimensional random variables for simplicity.

In the graph above,  the general correlation matrix  $\mathbf{K_x}$  of the two-dimensional random variable   $\mathbf{x} = (x_1, x_2)^{\rm T}$   is given,  where  $\sigma_1^2$  and  $\sigma_2^2$  describe the variances of the individual components.   $\rho$  denotes the correlation coefficient between the two components.

The random variables   $\mathbf{y}$   and   $\mathbf{z}$   give two special cases of  $\mathbf{x}$  whose process parameters are to be determined from the correlation matrices   $\mathbf{K_y}$   and   $\mathbf{K_z}$   respectively.

Hints:

• The exercise belongs to the chapter  Generalization to N-Dimensional Random Variables.
• Some basics on the application of vectors and matrices can be found on the pages   Determinant of a Matrix   and   Inverse of a Matrix .
• According to the page  "Contour lines for correlated random variables"  the angle $\alpha$ between the old and the new system is given by the following equation:

$$\alpha = {1}/{2}\cdot \arctan (2 \cdot\rho \cdot \frac{\sigma_1\cdot\sigma_2}{\sigma_1^2 -\sigma_2^2}).$$

• In particular,  note:
  • A  $2×2$-covariance matrix has two real eigenvalues  $\lambda_1$  and  $\lambda_2$.
  • These two eigenvalues determine two eigenvectors  $\xi_1$  and  $\xi_2$.
  • These span a new coordinate system in the direction of the principal axes of the old system.

Questions

1

Which statements are true for the correlation matrix  $\mathbf{K_y}$ ?

	$\mathbf{K_y}$  describes all possible two-dimensional random variables with  $\sigma_1 = \sigma_2 = \sigma$.
	The value range of the parameter  $\rho$   is  $-1 \le \rho \le +1$.
	The value range of the parameter  $\rho$   is  $0 < \rho < 1$.

2

Calculate the eigenvalues of  $\mathbf{K_y}$  under the condition  $\sigma = 1$  and  $\rho = 0$.

$\lambda_1 \ = \ $ $\ (\lambda_1 \ge \lambda_2)$

$\lambda_2 \ = \ $ $\ (\lambda_2 \le \lambda_1)$

3

Give the eigenvalues of   $\mathbf{K_y}$   under the condition   $\sigma = 1$   and   $0 < \rho < 1$   What values result for  $\rho = 0.5 $,  assuming  $\lambda_1 \ge \lambda_2$?

$\lambda_1 \ = \ $ $\ (\lambda_1 \ge \lambda_2)$

$\lambda_2 \ = \ $ $\ (\lambda_2 \le \lambda_1)$

4

Calculate the corresponding eigenvectors  $\mathbf{\eta_1}$  and  $\mathbf{\eta_2}$.  Which of the following statements are true?

	$\mathbf{\eta_1}$  and  $\mathbf{\eta_2}$  lie in the direction of the ellipse main axes.
	The new coordinates are rotated by  $45^\circ$.
	The standard deviations with respect to the new system are  $\lambda_1$  and  $\lambda_2$.

5

What are the characteristics of the random variable  $\mathbf{z}$  specified by  $\mathbf{K_z}$?

$\sigma_1 = \ $

$\sigma_2 = \ $

$\rho = \ $

6

Calculate the eigenvalues  $\lambda_1$  and  $\lambda_2 \le \lambda_1$  of the correlation matrix  $\mathbf{K_z}$.

$\lambda_1 \ = \ $ $\ (\lambda_1 \ge \lambda_2)$

$\lambda_2 \ = \ $ $\ (\lambda_2 \le \lambda_1)$

7

By what angle  $\alpha$  is the new coordinate system  $(\mathbf{\zeta_1}, \ \mathbf{\zeta_2})$  rotated with respect to the original system  $(\mathbf{z_1}, \ \mathbf{z_2})$ ?

$\alpha \ = \ $ $\ \rm deg$

Solution

Solution

(1)  Correct are proposed solutions 1 and 2:

• $\mathbf{K_y}$  is indeed the most general correlation matrix of a 2D random variable with  $\sigma_1 = \sigma_2 = \sigma$.
• The parameter  $\rho$  specifies the correlation coefficient.  This can take all values between  $\pm 1$  including these marginal values.

(2)  In this case, the governing equation is:

$${\rm det}\left[ \begin{array}{cc} 1- \lambda & 0 \ 0 & 1- \lambda \end{array} \right] = 0 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} (1- \lambda)^2 = 0\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \hspace{0.15cm}\underline{\lambda_{1/2} =1}.$$

(3)  With positive  $\rho$  the governing equation of the eigenvalues is:

$$(1- \lambda)^2 -\rho^2 = 0\hspace{0.5cm}\Rightarrow \hspace{0.5cm}\lambda^2 - 2\lambda + 1 - \rho^2 = 0\hspace{0.5cm}\Rightarrow\hspace{0.5cm}\lambda_{1/2} =1 \pm \rho.$$

• For  $\rho= 0.5$  one gets  $\underline{\lambda_{1} =1.5}$  and  $\underline{\lambda_{2} =0.5}$.
• By the way, the equation holds in the whole domain of definition  $-1 \le \rho \le +1$.
• For  $\rho = 0$  is  $\lambda_1 = \lambda_2 = +1$    ⇒   see subtask  (2)'.
• For  $\rho = \pm 1$  this gives $\lambda_1 = 2$  and  $\lambda_2 = 0$.

(4)  The eigenvectors are obtained by substituting the eigenvalues  $\lambda_1$  and  $\lambda_2$  into the correlation matrix:

$$\left[ \begin{array}{cc} 1- (1+\rho) & \rho \\ \rho & 1- (1+\rho) \end{array} \right]\cdot{\boldsymbol{\eta_1}} = \left[ \begin{array}{cc} -\rho & \rho \\ \rho & -\rho \end{array} \right]\cdot \left[ \begin{array}{c} \eta_{11} \\ \eta_{12} \end{array} \right]=0$$

$$\Rightarrow\hspace{0.3cm}-\rho \cdot \eta_{11} + \rho \cdot \eta_{12} = 0\hspace{0.3cm}\Rightarrow\hspace{0.3cm}\eta_{11}= {\rm const} \cdot \eta_{12}\hspace{0.3cm}\Rightarrow\hspace{0.3cm}{\boldsymbol{\eta_1}}= {\rm const}\cdot \left[ \begin{array}{c} 1 \\ 1 \end{array} \right];$$

$$\left[ \begin{array}{cc} 1- (1-\rho) & \rho \\ \rho & 1- (1-\rho) \end{array} \right]\cdot{\boldsymbol{\eta_2}} = \left[ \begin{array}{cc} \rho & \rho \\ \rho & \rho \end{array} \right]\cdot \left[ \begin{array}{c} \eta_{21} \\ \eta_{22} \end{array} \right]=0$$

$$\Rightarrow\hspace{0.3cm}\rho \cdot \eta_{21} + \rho \cdot \eta_{22} = 0\hspace{0.3cm}\Rightarrow\hspace{0.3cm}\eta_{21}= -{\rm const} \cdot \eta_{22}\hspace{0.3cm}\Rightarrow\hspace{0.3cm}{\boldsymbol{\eta_2}}= {\rm const}\cdot \left[ \begin{array}{c} -1 \\ 1 \end{array} \right].$$

To rotate the coordinate system

Putting this into what is called orthonormal form, the following holds:

$${\boldsymbol{\eta_1}}= \frac{1}{\sqrt{2}}\cdot \left[ \begin{array}{c} 1 \\ 1 \end{array} \right],\hspace{0.5cm} {\boldsymbol{\eta_2}}= \frac{1}{\sqrt{2}}\cdot \left[ \begin{array}{c} -1 \\ 1 \end{array} \right].$$

The sketch illustrates the result:

• The coordinate system defined by  $\mathbf{\eta_1}$  and  $\mathbf{\eta_2}$  is actually in the direction of the principal axes of the original system.
• With  $\sigma_1 = \sigma_2$  almost always results  $($exception:   $\rho= 0)$  the angle of rotation  $\alpha = 45^\circ$.
• This also follows from the equation given in the theory section:

$$\alpha = {1}/{2}\cdot \arctan (2 \cdot\rho \cdot \frac{\sigma_1\cdot\sigma_2}{\sigma_1^2 -\sigma_2^2})= {1}/{2}\cdot \arctan (\infty)\hspace{0.3cm}\rightarrow\hspace{0.3cm}\alpha = 45^\circ.$$

• The eigenvalues  $\lambda_1$  and  $\lambda_2$  do not denote the scatter with respect to the new axes, but the variances.

Thus, correct are the proposed solutions 1 and 2.

(5)  By comparing the matrices  $\mathbf{K_x}$  and  $\mathbf{K_z}$  we get.

• $\sigma_{1}\hspace{0.15cm}\underline{ =2}$,
• $\sigma_{2}\hspace{0.15cm}\underline{ =1}$,
• $\rho = 2/(\sigma_{1} \cdot \sigma_{2})\hspace{0.15cm}\underline{ =1}$.

(6)  According to the now familiar scheme:

$$(4- \lambda) \cdot (1- \lambda) -4 = 0\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\lambda^2 - 5\lambda = 0\hspace{0.3cm}\Rightarrow\hspace{0.3cm}\hspace{0.15cm}\underline{\lambda_{1} =5,\hspace{0.1cm} \lambda_{2} =0}.$$

(7)  According to the equation given on the specification sheet:

$$\alpha ={1}/{2}\cdot \arctan (2 \cdot 1 \cdot \frac{2 \cdot 1}{2^2 -1^2})= {1}/{2}\cdot \arctan ({4}/{3}) = 26.56^\circ.$$

Best possible decorrelation

The same result is obtained using the eigenvector:

$$\left[ \begin{array}{cc} 4-5 & 2 \\ 2 & 1-5 \end{array} \right]\cdot \left[ \begin{array}{c} \zeta_{11} \\ \zeta_{12} \end{array} \right]=0 \hspace{0.3cm} \Rightarrow\hspace{0.3cm}-\zeta_{11}= 2\zeta_{12}=0\hspace{0.3cm}\Rightarrow\hspace{0.3cm}\zeta_{12}={\zeta_{11}}/{2}$$

$$\Rightarrow\hspace{0.3cm}\alpha = \arctan ({\zeta_{12}}/{\zeta_{11}}) = \arctan(0.5) \hspace{0.15cm}\underline{= 26.56^\circ}.$$

The accompanying sketch shows the joint PDF of the random variable  $\mathbf{z}$:

• Because  $\rho = 1$  all values lie on the correlation line with coordinates  $z_1$  and  $z_2 = z_1/2$.
• By rotating by the angle  $\alpha = \arctan(0.5) = 26.56^\circ$  a new coordinate system is formed.
• The variance along the axis  $\mathbf{\zeta_1}$ is  $\lambda_1 = 5$  $($scatter  $\sigma_1 = \sqrt{5} = 2.236)$,
• while in the direction orthogonal to it  $\mathbf{\zeta_2}$  the random variable is not extended  $(\lambda_2 = \sigma_2 = 0)$.