Page 148 - Textos de Matemática Vol. 44

138 T. MORAN AND I. SPITKOVSKY
For singular A the choice of U is not unique; ∆(A) is nevertheless still defined by A uniquely. The transformation A → ∆(A) was introduced by Aluthge [1], and has since been studied extensively.
6.1. Unitary reducibility of ∆(A). If A is unitarily reducible, then the Aluthge transform acts on each block independently, so unitary reducibility is preserved under ∆. On the other hand, ∆(A) may be unitarily reducible even when A is not. In particular, the kernel of a singular matrix A is a reducing subspace for ∆(A). The criterion for ∆(A) to be normal (and thus unitarily reducible) in case of invertible A was obtained in [11, Theorems 8 and 9], stated in terms of either the polar representation of A or its unitary similarity to a certain canonical form. In the case of almost normal matrices, we propose here an alternative treatment which yields a more constructive result.
The following technical lemma is paramount.
Lemma 6.1. Let A ∈ Cn×n be an invertible pure almost normal matrix, n ≥ 3.
Then ∆(A) has no normal eigenvectors.
Proof. Suppose that ∆(A) and ∆(A)∗ have a common eigenvector x:
∆(A)x = ζx, ∆(A)∗x = ζx.
Using (6.1), these equalities can be equivalently rewritten in terms of A and R: R1/2AR−1/2x = ζx, R−1/2A∗R1/2x = ζx.
Denoting R−1/2x = y and R1/2x = z, we conclude from here that Ay = ζy, A∗z = ζz. In other words, y is an eigenvector of A and z is an eigenvector of A∗ corresponding to complex conjugate eigenvalues. It is crucial for the rest of the proof that
(6.2) z = R1/2x = R1/2(R1/2y) = Ry.
We now consider separately two situations, depending on which eigenvalue of A plays the role of ζ.
Case 1. ζ = λj for some j. Without loss of generality then y is ej, and z is a scalar multiple of the vector (3.2). Due to (6.2), z actually is the jth column of R. Since R is positive definite, its diagonal entries are positive, so μ must be different from λj. Moreover,
z=c[0 ... 010 ... 0βj/λj −μ]T
for some c > 0.
Consequently, in the notation R = (rik)ni,k=1 we have
rij =rji =0fori̸=j,n.