• Support PF! Buy your school textbooks, materials and every day products Here!

Positive definite quadratic forms proof

  • Thread starter wakko101
  • Start date
  • #1
68
0

Homework Statement


Given a real symmetric matrix A, prove that:

a) A is positive definite if and only if A = (B^T)B for some real invertible matrix B
b) A is positive semidefinite if and only if there exists a (possibly singular) real matrix Q such that A = (Q^T)Q

Homework Equations


quadratic form q(x) = a1*x1^2 + ... + an*xn^2
And possibly the principal axis theorem.

The Attempt at a Solution


for part a) I think I know how to show that A is pos def if you assume A = (B^T)B:
(P^T)AP = D = (P^T)(B^T)BP = ((PB)^T)PB which implies the diagonal entries of the diagonal matrix D are positive since row of (A^T) = column of A.
I'm not sure how to do it the other way, however. Any hints or advice would be appreciated.

Cheers,
W. =)
 

Answers and Replies

  • #2
jbunniii
Science Advisor
Homework Helper
Insights Author
Gold Member
3,394
180
For the "only if" part of (a), do you know a theorem that allows you to express any symmetric matrix in the form

[tex]A = U^T D U[/tex]

where [itex]U[/itex] is a unitary matrix (in fact, orthogonal since [itex]A[/itex] is real in your case) and [itex]D[/itex] is a diagonal matrix? If [itex]A[/itex] is positive definite, what can you say about the elements of [itex]D[/itex]?

Also, I'm not sure I buy your proof of the "if" part. A matrix [itex]A[/itex] is positive definite if for every real vector [itex]x[/itex],

[tex]x^T A x \geq 0[/tex]

If [itex]A = B^T B[/itex], then what does that imply about [itex]x^T A x[/itex]?
 
  • #3
68
0
I'm not entirely sure this is correct, but I've come up with the following:

If (X^T)AX > 0 then we have (Y^T)DY > 0 (for X = PY and D = (P^T)AP). This implies that the entries on the diagonal matrix D (the eigenvalues) are all positive. So, write D = (Q^T)Q where Q is the diagonal matrix of the square roots of D's elements. Then
D = (Q^T)Q = (P^T)AP = (P^1)AP (since P is orthogonal)
then P(Q^T)Q(P^1) = A
then (PQ)(Q^T)(P^T) (since (Q^T) = Q)
then (PQ)(PQ)^T = A
Take B = (PQ)^T and we have the result.

As for the other part, I don't think I quite finished my reasoning before....if we can show that the assumption requires that D have all positive entries (take them to be a1 to an) then q(v) = a1*xa^2 + ... + an*xn^2 will necessarily be positive. And since D = (PB)^T(PB) then they must be positive because the diagonal entries of D (which are the only ones that are not 0) are the rows of the transpose of PB multiplied by the columns of PB and these are the same vectors.

Cheers,
W. =)
 

Related Threads on Positive definite quadratic forms proof

Replies
0
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
1K
Top