Page 87 - Textos de Matemática Vol. 44

SOLUTIONS TO SESQUILINEAR MATRIX EQUATIONS 77
Proof. The canonical form of A is the direct sum of several Jordan blocks Jnk (μk) corresponding to nonnegative scalars μk. Now, the required assertion is obtained by combining Proposition 2.5 with the following simple observation: if L1 and L2 are coninvariant subspaces of A, then their sum L = L1 + L2 is an A-coninvariant subspace as well. Indeed,
AL=AL1 +AL2 ⊂L1 +L2 =L.

Proposition 2.7. Assume that the canonical form of A ∈ Mn(C) is the single block of type (2.7) corresponding to a nonreal scalar μ. (Thus, n = 2l is an even integer.) Then A has a coninvariant subspace of any even dimension from 2 to n.
Proof. It is easy to check that, for any k (1 ≤ k ≤ l), the 2k-dimensional subspace
L2k = span(e1,...,ek,el+1 ...,el+k)
is coninvariant with respect to G(μ). Then, by Proposition 2.3, PL2k is a 2k-
dimensional coninvariant subspace of A.
Corollary 2.8. Assume that no coneigenvalue of A ∈ Mn(C) is real. (Thus, n is necessarily an even integer.) Then A has a coninvariant subspace of any even dimension from 2 to n, and no coninvariant subspace of A has an odd dimension.
Proof. The first assertion of this corollary is justified as in the proof of Corol- lary 2.6. Suppose that an A-coninvariant subspace L has an odd dimension m. Let f1,...,fm be a basis in L. If
F = (f1 ···fm),
then
(2.8) AF = F B
for some m × m matrix B. Equality (2.8) implies that
ALF =FBL.
Since m is odd, BL must have a nonnegative eigenvalue, which gives rise to a nonnegative coneigenvalue of A. This contradiction proves the second assertion.
Theorem 2.9. If at least one coneigenvalue of A ∈ Mn(C) is real, then A has a coninvariant subspace of any dimension from 1 to n.