Math 2040 Chapter 7 Quadratic Forms and Symmetric Matricies 7.1 ...

Math 2040 Chapter 7 Quadratic Forms and Symmetric Matricies
7.1. Diagonalization of Symmetric Matrices
1. If A is a matrix with (i, j) th entry a i,j then A T denotes its transpose with (j, i) th entry a i,j . We
say A is symmetric if A T = A.
2. Suppose A is diagonalizable, so that A = P DP −1 where D is an n × n diagonal matrix.
When the change of coordinate matrix P is orthogonal then the A is said to be orthogonally
diagonalizable. Recall, P is orthogonal iff its columns are pairwise orthogonal unit vectors; in
which case, P T = P −1 .
3. Theorem 1 of this chapter shows that the eigenspaces of a symmetric matrix are pairwise orthogonal,
and Theorem 2 tells us every n × n symmetric matrix is orthogonally diagonalizable.
4. If A is symmetric and A = QDQ −1 with D diagonal, the Gram-Schmidt process applied to
Q followed by a normalization of each column constructs an orthogonal matrix P such that
A = P DP −1 = P DP T . This process will succeed in the orthogonal diagonalization of every
symmetric matrix. More exactly, to orthogonally diagonalize a symmetric matrix A follow these
steps:
(a) Find any old diagonalization A = QDQ −1 .
(b) If q k (1), ..., q k (r) are the columns of Q forming a basis of the eigenspace of λ k (in which
case λ k has multiplicity r) let s k (1), ..., s k (r) be the result of applying the Gram-Schmidt
process to q k (1), ..., q k (r). Let u k (i) = s k (i)/||s k (i)||, i = 1, ..., r. Then each u k (i) is a unit
eigenvector corresponding to λ k and these new eigenvectors are a basis of the eigenspace of
λ k .
(c) Form P by using the u k (i)’s found in the last step as columns for P . Then P is an orthogonal
matrix and A = P DP −1 as required.
⎡
5. Typical problem. Orthogonally diagonalize A = ⎣ 3 −2 4
⎤
−2 6 2 ⎦.
4 2 3
7.2. Quadratic forms
1. A Quadratic Form is an expression of the form Q(x) = x T Ax where A is a symmetric matrix
and x varies throughout R n . Obviously, Q(x) is a function R n −→ R.
[ ]
1 5
2. Example. If A = and Q(x) = x
5 1
T Ax then Q(x) = x 2 1 + 10x 1 x 2 + x 2 2.
3. Principal Axis Theorem. If Q(x) is a quadratic form then there is a change of coordinate
transformation x = P y such that Q(P y) does not contain any cross product terms.
Proof: Follow the steps below to find a change of coordinate matrix to eliminate the cross
product terms from quadratic form Q(x) = x T Ax.
(a) Orthogonally diagonalize A, so that A = P DP T for D a diagonal matrix and P an othhogonal
matrix.
(b) Change coordinates by setting x = P y (equivalently, y = P T x).
(c) The new quadratic form Q(P y) = x T Ax = x T (P DP T )x = (P T x) T D(P T x) = y T Dy will
not contain cross product terms. Indeed, Q(P y) = d 11 y 2 1 + d 22 y 2 2 + ... + d nn y 2 n.
4. Typical problem 7.2.7. Make a change of variable x = P y that transforms the quadratic form
Q(x) = x 2 1 + 10x 1 x 2 + x 2 2 into a quadratic form with no cross product terms. Give P and the
new quadratic form.

7.3. Constrained Optimization
1. The unit sphere in R n is S n−1 = {x | ||x|| = 1}. ||x|| 2 = x T x, so S n−1 = {x | x T x = 1}.
2. Theorem 6. (page 465 and 466) Q(x) = x T Ax be a quadratic form such that A is a symmetric
n × n matrix. Let M and m be the respective maximum (largest) and minimum (least, possibly
negative) values of Q(x) on S n−1 . Then M and m are the maximum and minimum eigenvalues
of A, respectively. Q(x) = M (respectively, Q(x) = m) at a unit eigenvector corresponding to
M (respectively, at a unit eigenvector corresponding to m).
3. Typical problem. Find the maximum value of Q(x) = x 2 1 + 10x 1 x 2 + x 2 2 subject to the constraint
x 2 1 + x 2 2 = 1.