Special Property of Some Symmetric Matrices 

In solving Ax = b it is necessary to employ special splitting properties followied by interation techniques to find the solution.  When the coefficient matrix is very large or nearly singular then there can be problems with Gauss elimination.  This particular property is best left to a numerical analysis class.  The property that will guarentee that the method works though is a property that we will study at this time.   

Definition An nxn matrix A is said to be postive definite if given any non-zero nx1 vector x then the product  

Let us look at the relationship of this property to eigenvalues.  Let be an eigenvalue of A with associated eigenvector .  Then we now that and we can look at the product .  Multiplying both sides of by we obtain 

.  If A is positive definite then we see that >0 .  Since  

is the length of squared then must be postive.  It is easy to show that if the matrix is symmetric then the eigenvalues being positive will make the matrix positive definite.  This follows from the fact that every symmetric matrix is similar to a diagonal matrix.  i.e. 

where Q is an orthogonal matrix and is a diagonal matrix with the diagonal elements on the diagonal.   Thus, and setting the right hand side becomes which is nothing more than + ... > 0 .   

There is a third way to check to see if a matrix is positive difinite.  That is that the determinate of all the principle submatrices are positive including the determinate of the matrix itself.   It will be best to illustrate what is ment by the principle submatrices.   

Let us now look at a few examples.  We will first show they are positive definite by using the definition and next using  

Example 1 

>	with(LinearAlgebra):

 

>

 

>	A:=<<4,1>|<1,3>>;

 

>

X:=<x,y>;

 

>	AX:=simplify(X^%T.A.X);

 

We now can write the prolynomial as =  

At this point it is obvious that every term is non-negative since that are all squared.  The remaining question are they positive.  The answer is yes since the only way they could be zero is if each term is zero but   tellus that x=0 and tell us that y=0.  The conclusion is that for any non zero vector the quadratic form is positive.  You need to get the quadratic expression as the sum of perfect squares to see that the quadratic form is positive.   

Let us look at the eigenvalues 

>	EA:=Eigenvectors(A,output='list');

 

>	evalf(EA[1][1]);

 

>	evalf(EA[2][1]);

 

What is are the pricipal submatrices of the matrix.  In this case there is only one and that one  

is the number in the first row and first column which is positive 

now for the det(A)= 12 -1 =11>0 so we have shown that the matrix is positive definite in three ways.    We can also look at the piviots after Gauss Elimination. 

>	GaussianElimination(A);

 

(2.1)

 

Since the piviots are positive then it is positive definite. 

The element in the first row and first column is positive .  Next look at the Determinant of the matrix.  This will be the principle show that all the principle subdeterminants are positive.  This is still another way of testing for positive definite. 

>	Determinant(A);

 

(2.2)

 

Exmaple 2 

Consider the following matrix.  

>	B:=<<1,1>|<1,1>>;

 

>	BX:=simplify(X^%T.B.X);

 

which tells us that and this is not always positive since for any x = -y we get zero.  i.e. for the vector [1,-1]. 

>	Eigenvectors(B,output='list');

 

>	Determinant(B);

 

>	GaussianElimination(B);

 

All the tests show that the matrix has non-negative eigenvalues put not position eigenvalues so it is not positive definite.  It is called positive semi-definite but we will not worry about pisitive semi-definite matrices in this class 

Example 3 

Another example 

>	C:=<<4,2>|<2,3>>;

 

>	CX:=simplify(X^%T.C.X);

 

Now we need to look at how to group our terms   so we arrive at the following factorization of = .  Again, the only way that this quanitity can be zero is if  you have the zero vector [0,0].  Why?  Let us check the other ways of looking at C being positive definite 

>	evalf(Eigenvectors(C,output='list'));

 

(4.1)

 

The 1,1 position is 4 so it is positive. 

>	Determinant(C);

 

(4.2)

 

Combining the 1,1 position being 4 with a positive determinant being positive tells us that the matrix is positive definite.  

>	GaussianElimination(C);

 

(4.3)

 

The piviots after Gauss Elimination being positive tells us in another way that the matrix is positive definite 

Example 4  

We will now look at a 3x3 matrix 

>	F:=<<4,1,0>|<1,4,1>|<0,1,4>>;

 

>	X1:=<x,y,z>;

 

>	X1T:=X1^%T;

 

>	X1F:=simplify(X1T.F.X1);

 

We need now to start collecting terms and finding perfect squares for our factors. 

=  

.  There are three terms which guarentee us that the only way for this quantity to be zero is for x=0,y=0 and z=0. i.e. we get the zero vector [0,0,0], so the matrix is positive definite by definition. 

>	evalf(Eigenvectors(F,output='list'));

 

This means that the eigenvalues are all postive. 

How about the principle submatries. 

> 0  

>	SF:=SubMatrix(F,1..2,1..2);

 

>	Determinant(SF);

 

>	Determinant(F);

 

>

 

We have thus shown that the matrix is positive definite in three different ways.   

Exercises 3.5  

Exercise 3.5.1 (a),(c),(d),(f).  (Exercise 3.5.6  (b) (d) You can use any of the Maple commands to work this last problem. ) You must follow the instructions  for exercise 3.5.1 Problems Due on  November 19, 2008 

e-mail me your worksheet.