A Householder matrix is symmetric

Question

I want to show that a Householder matrix is symmetric, so I must show that $H^T = H$, but from the formula

$$H= I - (uu^T/\beta),$$

they are not equal. What's wrong with my reasoning?

EDIT: I forgot that $(uu^T)^T$ would be $(u^T)^T(u)^T$ from the following properties: $(AB)^T=B^TA^T$

@VedranŠego: Thank you for your informative comment, I edited my question accordingly. — Gigili, Jan 06 '14 at 02:51

arbitUser1401 · Accepted Answer · 2014-01-06T06:19:13.383

8

Hint : $I$ is symmetric and $uu^{T}$ is symmetric.

Thus it follows,

$I^{T} =I$ and $(uu^{T})^{T}=uu^{T}$ (Recall that $(AB)^{T}=B^{T}A^{T}$ )

Hence $H^{T}=(I-uu^{T}/\beta)^{T}=I^{T}-(uu^T/\beta)^{T}=I-uu^{T}/\beta=H$

Verily , $H $ is symmetric.

edited Jan 06 '14 at 06:19

answered Jan 05 '14 at 20:32

arbitUser1401

Could you please do it by calculating the transpose matrix and showing how it is equal to $H$? – Gigili Jan 05 '14 at 20:43
$I^{T}=I$ and $(uu^{T})^{T}=uu^{T}$ thus $H^{T}=(I-uu^{T}/\beta)^{T}=I^{T}-(uu^{T}/\beta)^{T}=I-uu^{T}/\beta=H$ – arbitUser1401 Jan 05 '14 at 21:56
Thank you for your answer. It would be great if you could add your comment to your answer, because your answer is a bit incomplete as it is. – Gigili Jan 06 '14 at 02:49

score 2 · Answer 2 · answered Jan 05 '14 at 21:49

2

$$H = I - \frac {2} {u^T u} u u^T=I+\alpha \ u u^T$$

$$H^T=(I+\alpha \ u u^T)^T=I^T+\alpha \ (uu^T)^T=I+\alpha (u^T)^Tu^T=I+\alpha \ u u^T=H$$

Where I used $I^T=I$ and the basic properties of the transposed matrix, namely:

answered Jan 05 '14 at 21:49

Listing

2 Answers2