Thread: Jordan Decomposition to Schur Decomposition

Let A be a complex or real square matrix. Suppose we have a Jordan decomposition A = , where X is non-singular and J is upper bidiagonal. Show how you can obtain a Schur Decomposition from a Jordan Decomposition.


Schur Decomposition: A = QTQ*, where Q is unitary/orthogonal and T is upper triangular with the eigenvalues of A on the diagonal.


I'm really not sure at all what to do. Because Q is orthogonal, Q*= . I'm not sure if that plays in somehow.

I've been trying to use SVDs or QR decompositions of X or J to get there, but I've had no luck. Does anyone have any suggestions? Thank you so much.

QR should be right?

yields but are all upper triangular, so their product is as well, giving the required decomposition. The eigenvalue thing is a consequence of properties of triangular matrices and similarity preserving eigenvalues.