There is no such thing. Triangularization is a decomposition of a square matrix M into where T is a triangular matrix and P is some invertible matrix. It is not a decomposition of a linear operator (i.e. a linear map from a vector space to itself) because there is no notion of a triangular linear map. There is a notion of "triangularizability" of linear operators, however. That refers to whether it is possible to choose a single basis for the domain and codomain of the map such that its matrix representation in that basis is a triangular matrix. It's probably a bad choice of nomenclature, though, as a triangularizable linear operator cannot itself be triangularized (as I said, there is no such thing as a triangular map), but any of its matrix representations can.