## Transcript — Introduction

In this lecture, we will continue our look at quadratic forms. In particular, we will see how to classify quadratic forms and their corresponding symmetric matrices.

Definition: Let Q(x) be a quadratic form. We say that Q(x) is positive definite if Q is definitely positive—that is, if Q(x) is greater than 0 for all x not equal to the 0 vector. We say that Q(x) is negative definite if Q(x) is definitely negative—that is, Q(x) is less than 0 for all x not equal to the 0 vector. Q(x) is indefinite if Q(x) is greater than 0 for some x and Q(x) is less than 0 for some x. Q(x) is said to be positive semidefinite if Q(x) is greater than or equal to 0 for all x. And, Q(x) is said to be negative semidefinite if Q(x) is less than or equal to 0 for all x.

Let’s look at some simple examples. Example: Q(x1, x2) = x1-squared + x2-squared is positive definite since, clearly, Q(x1, x2) is greater than 0 for all x not equal to the 0 vector. Q(x1, x2) = -(x1-squared) – x2-squared is negative definite since Q(x1, x2) is clearly less than 0 for all x not equal to the 0 vector. Q(x1, x2) = x1-squared – x2-squared is indefinite since we can find a value of x—namely, [1; 0]—that gives a positive value, and another value of x—namely, [0; 1]—that gives a negative value. Q(x1, x2) = x1-squared is positive semidefinite since Q(x1, x2) is greater than or equal to 0 for all vectors x. Notice that it is equal to 0 for infinitely many different values of x—namely, all the values [0; x2].

We saw in the last lecture that quadratic forms corresponded directly with symmetric matrices, and so, not surprisingly, we classify symmetric matrices by classifying the associated quadratic form.

Example: Classify the symmetric matrix A = [1, 0; 0, -2]. Solution: The corresponding quadratic form is Q(x1, x2) = x1-squared – 2(x2-squared), and so we can see that Q(1, 0) = 1, and Q(0, 1) = -2, and so Q(x1, x2) is indefinite, and so A is also indefinite.

Example: Classify the quadratic form Q(x1, x2, x3, x4) = 7(x1-squared) – 2x1x2 + 2x1x3 – 6x1x4 + 7(x2-squared) – 6x2x3 + 2x2x4 + 7(x3-squared) – 2x3x4 + 7(x4-squared). Yikes! How are we going to classify this? In this form, it is very difficult to try to classify. It would be really nice if we could put this into an easier form. Wait—we can. We saw in the last lecture that orthogonally diagonalizing the corresponding symmetric matrix gives a change of variables that brings Q into diagonal form. And, as we saw in the previous examples, if Q is in diagonal form, then it is very easy to classify.

The symmetric matrix corresponding to Q is [7, -1, 1, -3; -1, 7, -3, 1; 1, -3, 7, -1; -3, 1, -1, 7]. With a little effort, we can find that the eigenvalues of A are lambda1 = 12, lambda2 = 8, lambda3 = 4, and lambda4 = 4. Hence, by Theorem 10.3.1, there exists an orthogonal matrix P such that the change of variables y = (P-transpose)x gives Q(x) = 12(y1-squared) + 8(y2-squared) + 4(y3-squared) + 4(y4-squared). This is clearly greater than 0 for all vectors y not equal to the 0 vector, and hence, is greater than 0 for all x not equal to the 0 vector since P is invertible. Consequently, Q and A are both positive definite.

Thus, it makes sense if we want to classify a quadratic form that we should diagonalize it. But when we diagonalize it, the coefficients are just going to be the eigenvalues of the associated symmetric matrix. And so, not surprisingly, we get the following theorem.

Theorem 10.3.2: If A is a symmetric matrix, then A is positive definite if and only if the eigenvalues of A are all positive, negative definite if and only if the eigenvalues of A are all negative, indefinite if and only if some of the eigenvalues of A are positive and some are negative, positive semidefinite if and only if the eigenvalues of A are all non-negative, and negative semidefinite if and only if the eigenvalues of A are all non-positive.

I will leave the proof as an easy exercise. To make the proof, you just need to generalize the method used in the last example.

## Examples

Example: Classify the following. (a) The quadratic form Q(x1, x2) = 4(x1-squared) – 6x1x2 + 2(x2-squared). Solution: We have the corresponding symmetric matrix is A = [4, -3; -3, 2], so the characteristic polynomial of A is the determinant of (A – (lambda)I), which equals lambda-squared – 6(lambda) – 1. Applying the quadratic formula, we get lambda equals 6 plus or minus the square root of 40 all divided by 2, so A has both positive and negative eigenvalues, and hence is indefinite. Thus, Q(x) is also indefinite.

(b) Q(x) is the quadratic form 2(x1-squared) + 2x1x2 + 2x1x3 + 2(x2-squared) + 2x2x3 + 2(x3-squared). Solution: The corresponding symmetric matrix is A = [2, 1, 1; 1, 2, 1; 1, 1, 2]. So, the eigenvalues of A are 1, 1, and 4. Hence, all of the eigenvalues of A are positive, and so A is positive definite. Therefore, Q(x) is also positive definite.

(c) The symmetric matrix A = [3, 2, 0; 2, 2, 2; 0, 2, 1]. Solution: We can find the eigenvalues of A are 5, 2, and -1, so A has both positive and negative eigenvalues, and hence, A is indefinite.

Notice that as long as you can quickly and correctly find the eigenvalues of a symmetric matrix, it is very, very easy to classify the symmetric matrix and its corresponding quadratic form.

In the next lecture, we will see a geometric interpretation in R2 of these classifications, as well as a geometric interpretation of much of our work with symmetric matrices.

This ends this lecture.