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P. 303
Unit 28: Positive Forms and More on Forms
Notes
= ( x x k ( f j , k )
j
j k
= ( A x x k ) ...(1)
kj j
j k
So we see that f is non-negative if and only if
and
A = A*
A x x 0 for all scalars x , x , ... x ..(2)
kj j k 1 2 n
j k
For positive f, the relation should be true for all (x , x , ... x ) 0. The above conditions on positive
1 2 n
f form are true if
g(X, Y) = Y*AX ...(3)
is a positive form on the space of n × 1 column matrices over the scalar field.
Theorem 1: Let F be the field of real number or the field of complex numbers. Let A be an n × n
matrix over F. The function g defined by
g(X, Y) = Y*AX ...(4)
n×1
is a positive form on the space F if and only if there exists an invertible n × n matrix P with
entries in F such that A = P*P.
Proof: For any n × n matrix A, the function g in (4) is a form on the space of column matrices. We
are trying to prove that g is positive if and only if A = P*P. First, suppose A = P*P. Then g is
Hermitian and
g(X, X) = X*P*PX
= (PX)*PX
0.
If P is invertible and X 0, then (PX)*PX > 0.
Now, suppose that g is a positive form on the space of column matrices. Then it is an inner
product and hence there exist column matrices Q , ..., Q such that
1 n
= g(Q , Q )
jk 1 k
= Q AQ .
*
k j
But this just says that, if Q is the matrix with columns Q , ..., Q , then A*AQ = I. Since {Q , ..., Q }
1 n 1 n
–1
is a basis, Q is invertible. Let P = Q and we have A = P*P.
In practice, it is not easy to verify that a given matrix A satisfies the criteria for positivity which
we have given thus far. One consequence of the last theorem is that if g is positive then det
A > 0, because det A = det (P*P) = det P* det P = |det P| . The fact that det A > 0 is by no means
2
sufficient to guarantee that g is positive; however, there are n determinants associated with A
which have this property: If A = A* and if each of those determinants is positive, then g is a
positive form.
Definition: Let A be an n × n matrix over field F. The principal minors of A are the scalars (A)
k
defined by
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