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A (real) symmetric, positive semidefinite (n × n) matrix V satisfies the condition

(4.196)

for any real (n × 1) vector a – irrespective of size or type of V, and of magnitude or sign of its elements, as long as the said product can be calculated; it should be emphasized that a^T Va represents a scalar. In the case of V being symmetric – and recalling the multiplication of any significant vector by a null matrix as per Eq. (4.67), one readily finds that

(4.197)

because Va =0_n×1 implies a^T 0_n×1 = 0 in general. To show the converse, one may resort to any form of column vector en lieu of a, namely, that obtained from addition of λb to a – with b denoting a vector of appropriate dimensions and λ denoting a scalar; a quadratic polynomial P{λ} may accordingly be defined as

(4.198)

while

(4.199)

in view of Eq. (4.196) – with no restriction imposed upon column vector a, or a + λb, for that matter. Algebraic expansion of Eq. (4.198) leads to

(4.200)

at the expense of Eqs. (4.24), (4.34), (4.76), (4.82), (4.114), and (4.123); as V is, by hypothesis, symmetric, one may resort to Eq. (4.195) to transform Eq. (4.200) to

(4.201)

also with the aid of the commutativity and associativity of addition of scalars. Since P{λ} cannot change sign as per Eq. (4.199) when λ spans the whole real domain, then its sign should remain that of the coefficient of λ² – as will be seen later, when dealing with the roots of a quadratic equation. In fact, absence of real distinct roots (as entertained by P{λ} ≥ 0 implies one of two possibilities: either a set of conjugate complex roots, say, r₁ = α + ιβ and r₂ = α − ιβ – so Eq. (2.182) indicates that P{λ} = a₂(λ − r₁)(λ − r₂), or else P{λ} = a₂(λ − (α + ιβ))(λ − (α − ιβ)) = a₂((λ − α) − ιβ)((λ − α) + ιβ), where the said product of two conjugate binomials reduces to P{λ} = a₂((λ − α)² − ι² β²), or P{λ} = a₂((λ − α)² + β²) since −ι² = 1, with (λ − α)² + β² > 0 for being the sum of two squares; or a double root r – in which case P{λ} = a₂(x − r)² is also based on Eq. (2.182). In either case, the factor multiplying a₂ being positive will be unable to change the sign brought about by a₂. Moreover, V is positive semidefinite by hypothesis, so a^T Va is nonnegative in agreement with Eq. (4.196) – and exchange of a with any other (compatible) vector b will also lead to b^T Vb ≥ 0, as expected. No distinct real roots can exist for P{λ}, so its discriminant binomial, Δ_P, must be negative or nil, i.e.

(4.202)

– where 4 was meanwhile factored out and the associative property of the multiplication of scalars duly utilized; if a^T Va = 0 as stated in Eq. (4.197), then Eq. (4.202) reduces to

(4.203)

that implies

(4.204)

because any other value of b^T Va would make the left‐hand side positive for it being a square. Equation (4.204) implies, in turn,

(4.205)

since no restriction was imposed on b; therefore, a^T Va = 0 implies Va =0_n×1, besides Va =0_n×1 implying a^T Va = 0, as seen previously – so Va =0_n×1 and a^T Va = 0 are equivalent statements. Consequently, Va ≠0_n×1 accounts for the inequality case in Eq. (4.196), i.e. a^T Va > 0.

If P denotes an (n × m) matrix, then the (m × m) matrix P^T VP is (real) symmetric, positive semidefinite; in fact,

(4.206)

as per Eqs. (4.110) and (4.120) – so P^T VP abides to the symmetry requirement because, by hypothesis, V^T = V, i.e.

(4.207)

Furthermore, the scalar

(4.208)

– where a and b denote (m × 1) and (n × 1) column vectors, respectively, and Eqs. (4.57) and (4.120) were taken advantage of, will be positive if b ≡ Pa ≠ 0_{n × 1} and nil if b ≡ Pa = 0_{n × 1}, in agreement with the foregoing derivation, coupled with Eqs. (4.196) and (4.197); this is so because V is positive semidefinite by hypothesis. In other words,

(4.209)

– which, together with Eq. (4.207), confirms the full original statement on P^T VP, i.e. a symmetric and positive definite matrix.

In the particular case of V = I_n, Eq. (4.196) degenerates to

(4.210)

at the expense of the n‐dimensional analogue of Eq. (3.8) – with the aid of Eq. (4.64) and explicitation of (n × 1) matrix a; hence, I_n is positive semidefinite as per Eq. (4.196). One may therefore resort directly to Eq. (4.209) to write

(4.211)

encompassing (m × 1) matrix a; and Eq. (4.211) may be algebraically simplified to

(4.212)

in view of Eq. (4.64). Inspection of Eq. (4.212) confirms that P^T P is necessarily a positive semidefinite matrix, while

(4.213)

as per Eq. (4.120), or else

(4.214)

in view of Eq. (4.110) – thus confirming its symmetry.