2

I am not familiar with this topic and have a naive question about QZ decomposition, which is defined as

For any matrix A and B in $\mathbb{R}^{n\times n}$, there exists orthogonal Q and Z, s.t. $QAZ=T$ upper-quasi triangular and $QBZ=S$ upper triangular.

My question is, could this mapping from $(A,B)$ to $(Q,QA,QB)$ unique (with some ordering and normalization)? Then is it continuous?

Any form of help would be appreciated, thank you so much!

Kryvtsov
  • 151
  • 1
    There is no mapping unless $Q$ and $Z$ are in fact unique. Have you considered basic examples (even with just a single matrix $A$) to explore this issue? – Ted Shifrin Aug 09 '22 at 18:12
  • Thank you for your comment! Like I said in the question I'm really not familiar with this topic. I barely know how to do a QZ decomposition and not to mention its properties. The reason I ask this is I have a program that does QZ decomposition (so it would always give me a unique result) in some part of it, and I would love to know if it is a continuous map. – Kryvtsov Aug 09 '22 at 18:35
  • I edited the question to be clearer, thanks! – Kryvtsov Aug 09 '22 at 18:58
  • You can see it's totally non-unique when $A$ is the identity matrix, for example, or has a sub-block that's the identity. If $B$ has a similar structure, then the $QZ$ decomposition is very non-unique ($Q$ can be arbitrary and $Z$ its transpose ... or the same of the appropriate blocks). – Ted Shifrin Aug 09 '22 at 19:02
  • That's right... Is there any way to confine the decomposition to be unique? I found that MATLAB always give a unique solution. I guess there must be some conventional `default' setting? – Kryvtsov Aug 09 '22 at 19:10
  • I have no experience with Matlab coding; sorry :) – Ted Shifrin Aug 09 '22 at 19:10
  • Thats fine! Thank you! I'm just surprised there does not exist a universally default way to do QZ, like Gaussian elimination when achieve a row-reduced form. I do realize the generalized eigenvalues are unique, not sure how that could help with the proof here. – Kryvtsov Aug 09 '22 at 19:17

0 Answers0