Page 88 - Textos de Matemática Vol. 44

78 KH. D. IKRAMOV
Proof. Assume that A has the real coneigenvalues μ1 , . . . , μr . (Some of them may be identical.) Then s = n − r coneigenvalues of A are nonreal. (Thus, s is an even integer.) According to Corollary 2.6, A has coninvariant subspaces
L1,L2,...,Lr
corresponding to the increasing subsets of the real conspectrum μ1, . . . , μr. By
Corollary 2.8, A has coninvariant subspaces M2,M4,...,Ms
of the indicated even dimensions that correspond to its nonreal coneigenvalues. The subspaces
Ni = M2i + L1, 1 ≤ i ≤ s − 1,
are again A-coninvariant and have the odd dimensions from 3 to s − 1. Finally,
the A-coninvariant subspaces
Pj =Ms+Lj, 1≤j≤r,
fill the remaining vacancies in the dimension sequence. Corollary 2.10. A matrix A of an even dimension m = 2n has no n-dimen-
sional coninvariant subspace only if n is odd and no coneigenvalue of A is real. 3. Sesquilinear matrix equations
We return to equation (1.2). Assume that it is solvable and X is an arbitrary solution to (1.2). We associate with X the m × n matrix
Define Then
MX =−B−DX.
−B −D In MX
In UX = X .
MUX= C A X=C+AX MX MX
= −XB−XDX = XM =UXMX. X
We conclude that the columns of UX constitute a basis in an n-dimensional coninvariant subspace of M.