Linear Algebra and Its Applications · Problem 6.2.36

Positive Definite Matrices — Gilbert Strang

Mathematics · tests-for-positive-definiteness

Problem

An algebraic proof of the law of inertia starts with the orthonormal eigenvectors x1,…,xpx_1,\ldots,x_p of AA corresponding to eigenvalues λi>0\lambda_i > 0, and the orthonormal eigenvectors y1,…,yqy_1,\ldots,y_q of CTACC^T AC corresponding to eigenvalues μi<0\mu_i < 0. (a) To prove that the p+qp+q vectors x1,…,xp, Cy1,…,Cyqx_1,\ldots,x_p,\, Cy_1,\ldots,Cy_q are independent, assume that some combination gives zero: a1x1+⋯+apxp=b1Cy1+⋯+bqCyq(=z, say).a_1 x_1 + \cdots + a_p x_p = b_1 C y_1 + \cdots + b_q C y_q \quad (= z,\text{ say}). Show that zTAz=λ1a12+⋯+λpap2≥0z^T A z = \lambda_1 a_1^2 + \cdots + \lambda_p a_p^2 \geq 0 and zTAz=μ1b12+⋯+μqbq2≤0z^T A z = \mu_1 b_1^2 + \cdots + \mu_q b_q^2 \leq 0. (b) Deduce that the aa's and bb's are zero (proving linear independence). From that deduce p+q≤np + q \leq n. (c) The same argument for the n−pn-p negative λ\lambda's and the n−qn-q positive μ\mu's gives n−p+n−q≤nn-p+n-q \leq n. (We again assume no zero eigenvalues---which are handled separately). Show that p+q=np+q=n, so the number pp of positive λ\lambda's equals the number n−qn-q of positive μ\mu's---which is the law of inertia.

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