On Symmetrizability of Hyperbolic Matrix Spaces

A new symmetrizability criterion for linear matrix spaces is proposed, with applications to the theory of first order conservation laws. Let L ⊂ Mat(n, k) be a real linear subspace of the space of (n × n)-matrices with coefficients from the field k = R or C. Definition 1. The family L is said to be hyperbolic if (1) A ∈ L for all A ∈ L, and all matrices in L have a simple real spectrum (i.e., the eigenvalues of any matrix A ∈ L are real and there is a basis consisting of the corresponding eigenvectors). Condition (1) implies the relation AB +BA = (A+B) −A −B ∈ L for all A,B ∈ L (this means that L is a special Jordan algebra). In particular, we can define linear operators SA on L by the rule SAB = AB +BA. If L is a hyperbolic space, then its extension {A + λE | λ ∈ R}, obtained by adding the unit matrix E, is also a hyperbolic space. The hyperbolicity condition admits the following reformulation. Proposition 1. A space L that satisfies (1) and contains the unit matrix E is hyperbolic if and only if the linear operators SA have a simple real spectrum in L for all A ∈ L. Proof. Let L be a hyperbolic space. Then the spectrum σ(A) of every matrix A ∈ L is simple and real. We consider the symmetric bilinear form (A,B) = TrAB. Then (A,A) = ∑ λ∈σ(A) λ 2 > 0 for A = 0. Therefore, the form (·, ·) is positive definite and determines a scalar multiplication on L. Direct verification shows that the operators SA are symmetric with respect to this scalar multiplication (i. e., (SAB,C) = (B,SAC) for all B,C ∈ L ); consequently, they have simple real spectra. Conversely, suppose that each operator SA has a simple real spectrum in L. Relation (1) and the condition E ∈ L imply that L contains all powers A, n ≥ 0, for A ∈ L, so that f(A) ∈ L for every real polynomial f(z). Let A ∈ L. By our assumptions, the operator SA has a simple real spectrum. Clearly, the simplicity of the spectrum of a matrix (or an operator) A means that there exists a polynomial p(z) = ∏m k=1(z − λk) with distinct real roots λk, k = 1, . . . ,m, such that p(A) = 0. Therefore, there exists a polynomial p(z) with distinct real roots such that p(SA) = 0. Since p(2A) = p(SA)E = 0 and the polynomial p(2z) also has distinct real roots, the spectrum of A is real and simple. The proof is complete. 2000 Mathematics Subject Classification. Primary 15A30, 15A06.