# Jordan Decomposition to Schur Decomposition

• Mar 29th 2009, 01:52 PM
azdang
Jordan Decomposition to Schur Decomposition
Let A be a complex or real square matrix. Suppose we have a Jordan decomposition A = $XJX^{-1}$, 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*= $Q^{-1}$. 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.
• Mar 30th 2009, 01:52 PM
siclar
QR should be right?

$X=QR$ yields $XJX^{-1}=QRJR^{-1}Q^{-1}=Q(RJR^{-1})Q^*$ but $R,J,R^{-1}$ 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.