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The concept of quadratic forms in linear algebra, focusing on the change of variable using a symmetric matrix p, the classification of quadratic forms as positive definite, negative definite, or indefinite, and the optimization of quadratic forms under certain constraints. The document also includes theorems related to the maximum and minimum values of quadratic forms.
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Differential Equations Math 54 Lec 005 (Dis 501) July 22, 2014
Given any basis B = {v 1 , · · · , vn} of Rn, let P =
(^) v 1 v 2 · · · vn
. Then,
y = P −^1 x
is the B-coordinate of x. So, P , kind of, changes a variable into another variable. Now, let A be a symmetric matrix and define a quadratic form xT^ Ax. Take P as the matrix of which columns are eigenvectors. (This can be done by the Spectral Theorem.) Especially, P has orthonormal column vectors. Hence,
xT^ Ax = yT^ (P T^ AP )y
and P T^ AP is a diagonal matrix of which diagonal entries are the eigenvalues of A. Such quadratic form can be written as
λ 1 y^21 + λ 2 y^22 + · · · + λny^2 n.
1.2.1 Positive Definite
Q(x) > 0 for all x 6 = 0
Furthermore, if Q(x) ≥ 0 for all x 6 = 0 , then Q is positive semidefinite.
1.2.2 Negative Definite
Q(x) < 0 for all x 6 = 0
Furthermore, if Q(x) ≤ 0 for all x 6 = 0 , then Q is negative semidefinite.
1.2.3 Indefinite
Q(x) has both positive and negative values.
In fact, the quadratic form 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 A has both positive and negative eigenvalues.
Furthermore, it is positive semidefinite if and only if the eigenvalues of A are nonnegative, negative semidefinite if and only if the eigenvalues of A are nonpositive.
Differential Equations Math 54 Lec 005 (Dis 501) July 22, 2014
Let A be a symmetric matrix that defines a quadratic form Q(x) = xT^ Ax. Then, under the condition that x is a unit vector, the maximum value of Q(x) is the largest eigenvalue λmax and the minimum value of Q(x) is the smallest eigenvalue λmin.
max{Q(x) : ||x|| = 1} = λmax, min{Q(x) : ||x|| = 1} = λmin
Under the same assumptions as 2.1, the maximum value of xT^ Ax subject to the constraints
xT^ x = 1, xT^ v 1 = 0
is the second largest eigenvalue λ 2.