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Unformatted text preview: ositive (or perhaps just nonnegative)? To answer this, let’s deal with real-symmetric matrices—the hermitian case follows along the same lines. If A ∈ n×n is symmetric, then, as
observed above, there is an orthogonal matrix P such that A = PDPT , where
D = diag (λ1 , λ2 , . . . , λn ) is real. If λi ≥ 0 for each i, then D1/2 exists, so
A = PDPT = PD1/2 D1/2 PT = BT B B = D1/2 PT , for D
E and λi > 0 for each i if and only if B is nonsingular. Conversely, if A can be
factored as A = BT B, then all eigenvalues of A are nonnegative because for
any eigenpair (λ, x), T
H 2 λ= Bx 2
xT BT Bx
2 ≥ 0.
R Moreover, if B is nonsingular, then N (B) = 0 =⇒ Bx = 0, so λ > 0. In other
words, a real-symmetric matrix A has nonnegative eigenvalues if and only if A
can be factored as A = BT B, and all eigenvalues are positive if and only if B
is nonsingular . A symmetric matrix A whose eigenvalues are positive is called
positive deﬁnite, and when the eigenvalues are just nonnegative, A is said to
be positive semideﬁnite.
The use of this terminology is consistent with that introduced in Example 3.10.7 (p. 154), where the term “positive deﬁnite” was used to designate
symmetric matrices possessing an LU factorization with positive pivots. It was
demonstrated in Example 3.10.7 that possessing positive pivots is equivalent to
the existence of a Cholesky factorization A = RT R, where R is upper triangular with positive diagonal entries. By the result of the previous paragraph this
means that all eigenvalues of a symmetric matrix A are positive if and only if
A has an LU factorization with positive pivots .
But the pivots are intimately related to the leading principal minor determinants. Recall from Exercise 6.1.16 (p. 474) that if Ak is the k th leading
principal submatrix of An×n , then the k th pivot is given by Y
C ukk = for k = 1,
det (A1 ) = a11
det (Ak )/det (Ak−1 ) for k = 2, 3, . . . , n. Consequently, a symmetric matrix is positive deﬁnite if and only if each of its
leading principal minors is positive . However, if each l...
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