Linearize the system about origin which is an equilibrium point.

Evaluating the coefficients of Eqn. (3), we get

$\displaystyle \frac{\partial f_1(k)}{\partial x_1(k)}=-1 + 3x_1^2(k) + x_2^2(k) \quad \frac{\partial f_1(k)}{\partial x_2(k)}=1 + 2x_1(k)x_2(k)$

Thus $A= \begin{bmatrix}-1 & 1 \\ -1 & -1 \end{bmatrix}$ . Hence, the linearized system around origin is given by

$\displaystyle \Delta \boldsymbol{x}(k+1) = \begin{bmatrix}-1 & 1 \\ -1 & -1 \end{bmatrix} \Delta \boldsymbol{x}(k)$

(9)

Sign definiteness of functions and matrices

Positive Definite Function: A continuously differentiable function $f:R^n\rightarrow R^+$ is said to be positive definite in a region $S \in R^n$ that contains the origin if

$ f(0)=0$

The function f (x) is said to be positive semi-definite

$ f(0)=0$

If the condition (2) becomes f (x) < 0 , the function is negative definite and if it becomes f (x) ≤ 0 it is negative semi-definite.

Example: Is the function positive definite?
Answer: f (0,0) = 0 shows that the first condition is satisfied. f (x₁,x₂) > 0 for x₁,x₂ ≠ 0.

Second condition is also satisfied. Hence the function is positive definite.

A square matrix P is symmetric if P = P^T. A scalar function has a quadratic form if it can be written as x^T Px where P = P^T and x is any real vector of dimension $n\times 1$ .

Positive Definite Matrix: A real symmetric matrix P is positive definite, i.e. P > 0 if

1. x^TPx > 0 for every non-zero x.

2. x^TPx = 0 only if x= 0.

A real symmetric matrix P is positive semi-definite, i.e.P ≥ 0 if x^TPx ≥ 0 for every non-zero x. This implies that x^TPx = 0 for some x ≠ 0 .

Theorem: A symmetric square matrix P is positive definite if and only if any one of the following conditions holds.

Every eigenvalue of P is positive.

All the leading principal minors of P are positive.

There exists an $n\times n$ non-singular matrix Q such that P = Q ^T Q .

Similarly, a matrix P is said to be negative definite if -P is positive definite. When none of these two conditions satisfies, the definiteness of the matrix cannot be calculated or in other words it is said to be sign indefinite.

Example: Consider the following third order matrices. Determine the sign definiteness of them. $\displaystyle A_1 = \begin{bmatrix}2 & 5 & 7\\ 1 & 3 & 4 \\ 1 & 2 & 5 \end{bmat... ...\;\;\; A_2 = \begin{bmatrix}2 & 0 & 0\\ 0 & 5 & -1 \\ 0 & 0 & -3 \end{bmatrix}$

The leading principal minors of the matrix A₁ are 2, 1 and 2, hence the matrix is positive definite.

The eigenvalues of the matrix A₂ can be straightaway computed as 2, 5 and -3, i.e., all the eigenvalues are not positive. Again, the eigenvalues of the matrix -A₂ are -2, -5 and 3 and hence the matrix A₂ is sign indefinite.