Another question brought this up. The only definition I have ever seen for a matrix being upper triangular is, written in component forms, “all the components below the main diagonal are zero.” But of course that property is basis dependent. It is not preserved under change of basis.
Yet it doesn’t seem as if it would be purely arbitrary because the product of upper triangular matrices is upper triangular, and so forth. It has closure. Is there some other sort of transformation besides a basis transformation that might be relevant here? It seems as if a set of matrices having this property should have some sort of invariants.
Is there some sort of isomorphism between the sets of upper triangular matrices in different bases?
Many true things can be said about upper-triangular matrices, obviously… 🙂
In my own experience, a useful more-functional (rather than notational) thing that can be said is that the subgroup of GLn consisting of upper-triangular matrices is the stabilizer of the flag (nested sequence) of subspaces consisting of the span of e1, the span of e1 and e2, … with standard basis vectors.
Concretely, this means the following. The matrix multiplication of a triangular matrix A and e1, Ae1, is equal to a multiple of e1, right? However, Ae2 is more than a multiple of e2: it can be any linear combination of e1 and e2. Generally, if you set Vi=span(e1,…,ei), try to show that A is upper triangular if and only if A(Vi)⊆Vi. The nested sequence of spaces
is called a flag of the total space.
One proves a lemma that any maximal chain of subspaces can be mapped to that “standard” chain by an element of GLn. In other words, no matter which basis you are using: being triangular is intrinsically to respect a flag with dim(Vi)=i (the last condition translate the maximality of the flag).
As Daniel Schepler aptly commented, while an ordered basis gives a maximal flag, a maximal flag does not quite specify a basis. There are more things that can be said about flags versus bases… unsurprisingly… 🙂