Let A be as in Problem and let Q(x) = x′Ax be the corresponding quadratic form. We call th...

Let A be as in Problem and let Q(x) = x′Ax be the corresponding quadratic form. We call this form positive definite if Q(x) > 0 for x ≠ 0. Show that, if Q(x) is positive definite, then det A > 0. [Hint: Use Problem 6 following Section 1.11.]

Remark It can be shown that Q is positive definite if and only if det Ak > 0 for k = I,... ,n, where Ak = (aij), i = 1,..., k, j = 1,..., k. See Chapter 6 of G. Strang, Linear Algebra and Its Applications, 3rd ed., Int’l Thomson Publishing, 1988.

Problem

Let A be a real symmetric matrix. Show, with the aid of the results of Problem, that all eigenvalues of A are real. [Hint: Let Av = λv for some complex λ and some nonzero complex vector v = col(v1, ..., vn). Consider the product Show that Q is real and that

Conclude that λ is real.]

Remarks By Problem, λ is actually an eigenvalue of A as a real matrix. The expression is a special case of a Hermitian quadratic form Q(v). See Section 5−7 of the book by Perlis listed at the end of the chapter.

Problem

We consider complex matrices, that is, matrices with complex entries. If A = (aij) is such a matrix, we denote by the conjugate of A, that is, the matrix , where z is the conjugate of the complex number z (for example, if z = 3 + 5i, then z = 3 – 5i).

Prove the following:

a) If z1 and z2 are complex numbers, then

b) If A and B are matrices and c is a complex scalar, then

c) Every complex matrix A can be written uniquely as A1 + iA2, where A1 and A2 are real matrices, and We call A1 the real part of A, A2 the imaginary par of A.

d) If A is a square matrix, then and, if A is nonsingular, then

e) A is a real matrix if and only if A =

Problem

Let A be a real square matrix, let λ be real, and let Av = λv for a nonzero complex column vector v. Show that Au = λu for a real nonzero vector u, so that λ is an eigenvalue of A considered as a real matrix. [Hint: Let v = p + iq, where p and q are real and not both zero. Show, with the aid of the results of Problem 10, that Ap = λp and Aq = λq and hence that u can be chosen as one of p, q.]