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: $(AB)^T=B^TA^T$

  • 1
    $\begingroup$ 1. These matrices are not named such because they are holding houses, but after Alston Scott Householder. 2. How can we tell you what are you doing wrong when you didn't write what were you doing? $\endgroup$ Jan 5, 2014 at 22:54
  • $\begingroup$ @VedranŠego: Thank you for your informative comment, I edited my question accordingly. $\endgroup$
    – Gigili
    Jan 6, 2014 at 2:51

2 Answers 2


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.

  • $\begingroup$ Could you please do it by calculating the transpose matrix and showing how it is equal to $H$? $\endgroup$
    – Gigili
    Jan 5, 2014 at 20:43
  • $\begingroup$ $I^{T}=I$ and $(uu^{T})^{T}=uu^{T}$ thus $H^{T}=(I-uu^{T}/\beta)^{T}=I^{T}-(uu^{T}/\beta)^{T}=I-uu^{T}/\beta=H$ $\endgroup$ Jan 5, 2014 at 21:56
  • $\begingroup$ 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. $\endgroup$
    – Gigili
    Jan 6, 2014 at 2:49

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

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

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

  • For scalars $\lambda$ we have $(\lambda A)^T=\lambda A^T$
  • $(A+B)^T=A^T+B^T$
  • $(AB)^T=B^TA^T$
  • $(A^T)^T=A$
  • $\begingroup$ Thank you, I figured where I was wrong. $\endgroup$
    – Gigili
    Jan 6, 2014 at 2:49

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