0

The Householder reflector $H = I - 2vv^T$ contains an outer product with $v$, which is not backwards stable. Why are Householder reflectors still used -- wouldn't that lack of stability cause an issue?

greg115
  • 463
  • 3
  • 11
  • Does this answer your question? https://math.stackexchange.com/questions/770882/why-is-householder-computationally-more-stable-than-modified-gram-schmidt – CyclotomicField Dec 09 '23 at 16:04
  • @CyclotomicField not exactly: I'm curious about Householder reflector usage in general, not specific to the QR decomp. But I found another source that answers my question; I'll write up an answer. – greg115 Dec 09 '23 at 16:06

1 Answers1

2

Indeed $vv^T$ is an outer product which is not numerically backwards stable. But for any vector $b$, we can compute $Hb$ with

$$Hb = (I - 2vv^T)b = b - 2(v^Tb)v$$

and since most uses of Householder reflectors will involve matrix-vector products, the issue of the outer product is sidestepped.

Source: 1

greg115
  • 463
  • 3
  • 11