I'm reading the book Matrix Analysis for Scientists and Engineers by Alan J. Laub. In the chapter about Canonical forms, the following theorem is presented:

Theorem 10.25 (Complete Orthogonal Decomposition): Let $A\in\mathbb{C}^{m\times n}_r$. Then there exist unitary matrices $U\in\mathbb{C}^{m\times m}$ and $V\in\mathbb{C}^{n\times n}$ such that $$UAV=\begin{bmatrix} R & 0\\ 0 & 0 \end{bmatrix} $$

where $R\in\mathbb{C}^{r\times r}_r$ is upper (or lower) triangular with positive diagonal elements.

Why is this relevant? It seems like I misunderstood this somehow, since to me this follows immediately from the SVD decomposition of $A$ (explained before in the book), and I expect $R$ to be diagonal, not only triangular. What am I missing?

  • 1
    $\begingroup$ Perhaps finding a triangular (rather than a diagonal) $A$ is computationally easier? Presumably such a decomposition is non-unique $\endgroup$ – Omnomnomnom Mar 1 at 14:34
  • 1
    $\begingroup$ @Omnomnomnom: I see your point. But then perhaps that makes the algorithm that produces this decomposition interesting, not the theorem itself. $\endgroup$ – JLagana Mar 1 at 14:49
  • 1
    $\begingroup$ I think that it is easy to immediately jump to SVD when you see $UAV=\begin{bmatrix} R & 0\\ 0 & 0 \end{bmatrix}$ since $U$ and $V$ are both unitary. However, SVD says there exists a factorization $A = URV^H$, where $U$ and $H$ are unitary, and $R$ is diagonal, so SVD makes the extra assumption that $R$ is diagonal, while the factorization you are asking makes the assumption that $R$ is triangular. I think it is just another matrix factorization that may or may not come in handy. $\endgroup$ – se2018 Mar 1 at 15:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.