Let [i.sub.A]: A [right arrow] T and [i.sub.C] : C [right arrow] T be the inclusion maps sending elements of A and B into the direct
summands A and C of T, respectively, i.e., [i.sub.A](a) = a [member of] A [subset] T and [i.sub.C](c) = c [member of] C [subset] T for every a [member of] A and c [member of] C.
The left question is how much
summands we need in the right-hand side.
We apply [[DELTA].sup.+.sub.2] to all
summands in the explicit formula for [[not upsilon].sub.2,[GAMMA]] (x).
If X = H is a Hilbert space and if T is a contraction, then T = C[direct sum]U is uniquely a direct sum of a completely nonunitary contraction C and a unitary operator U (where any of the these direct
summands may be missing) by Nagy-Foias-Langer decomposition for Hilbert-space contractions (see, e.g., [14, p.8] or [15, p.76]).
One decomposes this algebraic structure into
summands of the form 0 [right arrow] F [right arrow] ...
According to [8, 10], (11) has a unique solution if we set the
summandsOjah, Partitions with designated
summands in which all parts are odd, Integers, 15 (2015), #A9.
After the annihilation of equal
summands on both sides of obtained equality and after division by [a.sup.m.sub.0] = [c.sup.m] [b.sup.m.sub.0] using the of simple transformations, we can obtain the following relation:
Regarding the two
summands on the last line of (73), we have
and since the
summands are nonnegative it is evident that F(t) is SM.
Then,
summands [??] given by (6) with [[??].sup.m.sub.i](r) = 0 skipped and given by (8) and by (9) are equal to
Similarly, it is useful to classify together finite-dimensional modules over a finite-dimensional algebra with isomorphic indecomposable
summands - two such modules, which have the same indecomposables but perhaps with different nonzero multiplicities, are said to be similar.