Why don't we consider the zero subspace (which is readily $T$-invariant) in the definition of direct sum of linear operators? !enter image description here > Why don't we consider the zero subspace (which is readily $T$-invariant) in the definition of direct sum of linear operators? REF: Schaum's Outline of Linear Algebra

Technically, you could.

A decomposition of an operator is much like a factorization of a natural number. We can consider factorizations of numbers with '1' as one of the factor, but then we really are interested in is non-trivial factorizations.

I would prefer your definition, allowing for zero subspaces, but we don't lose much either way. I suspect that is why `(nonzero)` is in parentheses - if they felt it was a hard rule, they wouldn't need to parenthesize that, would they?

Essentially, we are interested in whether an operator is "decomposable." Non-decomposable operators are, in some sense, primes. But all operators are decomposables if we allow for the null space, so we would then need to define non-decomposable as "only has trivial decompositions."