Chapter 28.3

1E

Consider a set of unit vectors.

Where, and so on.

That is for every all vectors value are zeros, expect for row which is 1.

Symmetric Positive Definite Matrix: If a symmetric matrix satisfy the condition for all non-zero column vectors of real numbers, then the matrix is said to be a symmetric positive definite matrix.

Consider the symmetric positive definite matrix A as follows:

The condition holds for all .

Now suppose for any value in diagonal of matrix A is not positive, that is .

Choose from the set X and calculate.

That is . So, will be negative.

This is a contradiction to the initial consideration; A is a symmetric positive definite matrix.

Therefore, if A is a symmetric positiv e definite matrix then the value of , . That is all the diagonal elements of the symmetric positive definite matrix are positive.

2E

The matrix is which is symmetric positive definite matrix. Its determinant is positive by “completing the square”. It can be proved as follows:

Let x = -by/a and A is positive definite so,

=

The determinant of the above equation is given as:

Here, b = 2by, a = a, c =

So, substituting the values of a, b and c in D it can be deduced that:

Ignoring the constant value that is 4, the expression becomes:

=

Thus, so, it means that >0 because a>0

3E

A symmetric positive-definite matrix:

Let A is the matrix which is real in nature and can be positive-definite for a non-zero vector, if:

A symmetric positive-definite matrix has following properties:

• All the Eigenvalues of A will be positive.

• The largest element of the matrix is always there on the diagonal.

• All the diagonal elements of A will be positive.

• Matrix A will be a nonsingular matrix.

Consider the symmetric positive-definite matrix A as:

To prove that the maximum element lies on the diagonal, apply the same permutation to the both rows and columns of matrix A, the resultant matrix will continue to be symmetric and positive-definite.

The SYLVESTER’S CRITERION states that: A symmetric matrix A is positive-definite if and only if:

…… (1)

Where, A is the matrix formed by the intersection of the first k rows and columns of A and in a symmetric matrix .

If there is any element at position, it will also there at the position of. Thus, every element which is not at a diagonal, it will also appear at another position of the matrix. Further, for, applying the suitable permutation, SYLVESTER’S CRITERION gives:

…… (2)

Now, consider the positive-definite matrix given below:

In symmetric, real, positive-definite matrix, there is always large multiplier with a positive-definite matrix.

• In the above matrix, large multiplier is 10001. Due to this multiplier, the determinant of the matrix becomes positive.

• Knowing this, large multipliers will not try to create unstable conditions unless they lead to growth of the transformed element, because the length of matrix continuously increases.

• And to make the determinant positive, the diagonal elements products have to be made larger than the non-diagonal terms, which are used in the corresponding calculation.

• Further, as all the reduced matricesare positive-definite, their maximum element will lie on the diagonal (consider the above determinant equation), which is one of their properties stated above.

• Also from equation (2) either one of the element that is or should be greater than to hold the inequality of equation (1). Hence the max element should lie in the diagonal.

4E

A symmetric positive-definite matrix:

A matrix A, which is real in nature, can be positive-definite for a non-zero vector, if:

… … (1)

A symmetric positive-definite matrix has following properties:

a. All the Eigen values of A will be positive.

b. The largest element of the matrix is always there on the diagonal.

c. All the diagonal elements of A will be positive.

d. A will be a nonsingular matrix.

Consider a symmetric positive-definite matrix A as:

To prove that determinant of each leading sub-matrix of a symmetric positive matrix is positive, suppose that be any leading sub-matrix of a symmetric positive-definite matrix.

After removing corresponding rows and columns of the symmetric matrix, the sub-matrix is also going to be symmetric in nature.

So, is a symmetric matrix. … … (a)

Now, consider that be a particular column inside this sub-matrix, having p entries, which are non-zero in nature.

Increase to make it into an n-vector by inserting zeros into the places corresponding to the particular rows ofwhich were previously removed in the form. Then is not the zero vector and a relation is derive.