invariant complement
Given an invariant subspace W, an ordinary complement is any U with V = W (+) U — they share only zero and together span everything. An invariant complement is the demanding version: U is itself invariant under T. When W has an invariant complement, the pair (W, U) reduces T into two independent operators.
Why demand invariance of the complement? Because then, and only then, the matrix of T becomes block-diagonal instead of merely block-triangular. With a non-invariant complement you still get an upper block-triangular shape, but T leaks from U into W through the off-diagonal block; an invariant U seals that leak shut, so T = T|W joined with T|U with nothing in between.
Existence is not automatic. The classic failure is the shear [1, 1; 0, 1]: the x-axis is invariant, yet it has no invariant complement — any other invariant line would be a second eigendirection, and there is only one. An operator for which every invariant subspace has an invariant complement is exactly a semisimple operator, the structural meaning of complete reducibility.
An invariant subspace with no invariant complement — the hallmark of a non-semisimple (here purely Jordan) operator.
For normal operators (and self-adjoint or unitary ones) the orthogonal complement of an invariant subspace is always invariant too — that is why the spectral theorem gives a clean orthogonal block-diagonalization.