[Math] What’s the difference between Jordan and Schur decomposition

Question

They both seems to decompose a square matrix into a upper triangular matrix, but what's the fundamental difference between these two decompositions?

Thanks!

Answer 1

The Schur decomposition uses unitary transformations. This makes it reasonable from a numerical computation standpoint. It also gives an idea of how close the matrix is to being normal.

It can be basically impossible to compute the Jordan form numerically when eigenvalues are repeated and eigenvectors are 'close to being parallel'.

The Jordan form is often easier to work with from a theoretical standpoint (fewer numbers to deal with).