Diagonalizability w.r.t. non-orthonormal basis in application

Question

Are there applications of linear algebra where the diagonalizability with respect to non orthonormal basis is important? What I can find is usually that the operator is Hermitian and so we can assume an orthonormal basis in which the operator is diagonal, e.g., SVD of data (via symmetry of $A^TA$) or quantum mechanics (observables are Hermitian).

Answer 1

If $A$ is a square matrix, and you want to know what $A^n$ looks like for large $n$, then, as long as $A$ is diagonalizable, whether to an orthonormal basis or not, you can get $A^n=PD^nP^{-1}$ where $D$ is diagonal (and so $D^n$ is trivial to compute, and thus so is $A^n$).

Answer 2

In an euclidean or unitary space, non orthonormal bases consist of a dual pair $$(e_i),(e^k): e^k(e_i) = \delta^k_i.$$ The dual base is canonically implemented as the projector $$e^k(v^i e_k) = v^k$$ but even this simplifying fix can be generalized by the choice of any other basis in the dual space.

This implies that all elements of the tensor algebra have numerous representations in co- and contravariant components with respect to the two bases.

Inevitable in Riemannian geometries and the tensor algebra over its pair of dual bases.

In order not not mix up things like a physicist, in abstract linear algebra the space and its dual space of linear maps to the number field, aka component projector in a basis representation, are considered as distinct, but isomorphic spaces $X,L(X)$.

In infinite dimensional spaces witout a scalar product, the two spaces are very different, the smaller the vector space, the bigger its dual space, as can be seen if integer series are replaced by integrable functions over a real interval and sums over the integers are replaced by integrals.

With its scalar product, the exceptional class are Hilbert spaces, where again on recurses to consider the two spaces identical with the same basis.