Question: Suppose that $u$ is a vector in $\mathbb{R}^{3}$ of length 1, and let $w = u^{t}$ be the transpose of $u$. Let $R$ be the matrix $R = I - 2uw$. Explain why $R$ is called a reflection matrix, by explaining why, for any vector $v, wv = \langle u,v \rangle$, and explaining what $R$ does to the vectors on the line spanned by $u$, and what $R$ does to the vector perpendicular to $u$.

I've been spending a lot of time going over linear transformation and watching tutorials on transformation matrices. I found out that this matrix is actually called a 'householder reflection' but I couldn't find too much information about it in terms of the question asked.

I can prove that $R$ is orthogonal using $RR^{T}= I$. I'm not sure if that helps much.

I know if $\|\vec{u}\| = 1 $ then $uu^{T} = 1$

Because an orthogonal matrix preserves length and angles that means it limits the transformations to reflections and rotations.

Any vector that is perpendicular is left unchanged. Let's say $z$ is perpendicular to $u$ $$(I-2uu^{T})z = z-2u(u^{T}z)=z$$

I can't really get any further than this.

  • $\begingroup$ It's a "Householder reflection"; Householder is someone's name. $\endgroup$ – Omnomnomnom Nov 11 '16 at 12:28

As you rightly discovered, we find $Rz = z$ whenever $z\perp u$.

On the other hand, if $z$ is parallel to $u$, we have $z = \alpha u$ for some constant $\alpha$, so that $$ Rz = \alpha (Ru) = \alpha (I - 2uu^T)u = \alpha(u - 2u(u^Tu)) = \alpha(u - 2u) = -\alpha u = -z $$ where $u^Tu = 1$ since $u$ has length $1$. That is, $Rz = -z$; $R$ reflects the vector across the origin.

Since we know what $R$ does to vectors parallel to $u$ and those perpendicular, we can conclude that $R$ is the reflection along $u$ across the plane perpendicular to $u$.

  • 1
    $\begingroup$ That's so well worded. Thank you very much. $\endgroup$ – Patrick Moloney Nov 11 '16 at 12:36

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