The 2x2 case can be solved completely. The general 2x2 matrix can be written [a0+a3 a1-ia2] [a1+ia2 a0-a3 ] = M. M is Hermitian iff a0,...,a3 are real. M has eigenvalues a0 +/- sqrt(a1^2 + a2^2 + a3^2). The roots of a quadratic polynomial with integer coefficients are representable by a rational 2x2 Hermitian matrix iff the discriminant is the sum of three rational squares. [I'm abusing notation by calling a number rational if its real and imaginary parts are rational.] The polynomial x^2-7 is not so representable since 7 is not the sum of three rational squares. To see this, suppose 7 = (m1/d1)^2 + (m2/d2)^2 + (m3/d3)^2 = (n1^2 + n2^2 + n3^2)/d^2. Then 7d^2 is the sum of three integer squares. It's a theorem that all nonnegative integers are the sum of three squares, except those of the form 4^a (8b+7). If d is even, d^2 is a power of 4. If d is odd, d^2 == 1 mod 8, so 7d^2 == 7. Thus 7d^2 cannot be the sum of three integer squares. If you require not not only that M be Hermitian, but also real, then a2=0, and the discriminant must be the sum of two rational squares. -- Gene ________________________________ From: Henry Baker <hbaker1@pipeline.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Mon, November 16, 2009 6:16:25 AM Subject: [math-fun] characteristic polys of real symmetric matrices We would like to represent the polynomial t^2+2*b*t-c as the characteristic polynomial of a 2x2 matrix. By translating the origin, we can get rid of b, which makes the following analysis more perspicuous. So we have p(t)=t^2-c as the polynomial we want to represent by the symmetric real matrix [x y] [y z] = M. This matrix M satisfies its own characteristic polynomial, so we have the two equations: y*(x+z)=0 x^2+y^2-c=0 The first equation gives z=-x, while the second gives x=+- sqrt(c-y^2). So, M is [-sqrt(c-y^2) y] [y sqrt(c-y^2)] Thus, so long as |y|<sqrt(c), M will represent p(t). In general, for 2x2 matrices, y^2 enters as an addition to the discriminant, so we have some freedom to choose y if the discriminant is positive. Unfortunately, we have effectively solved the polynomial p(t) already, prior to constructing M, so it would appear that constructing M is not easier than solving p(t).