Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have trouble understanding the following exercise so I would really appreciate any help you could give me:

Let $k$ be a non zero vector in $\mathbb R^n$, written in standard basis. Let $H$ be defined as $H:=I-2\dfrac{\mathbf{n}\mathbf{n}^T}{\|\mathbf{n}\|^2}$, where $I$ is the identity matrix:

a) Prove, that $V = \{ u\in \mathbb{R}^n : k^Tu = 0\}$ is a vector subspace in $\mathbb{R}^n$ . What is its dimension?

b) Prove, that the transformation matrix $H$ mirrors over the subspace of $V$ in $\mathbb{R}^n$ (you have to prove, that $Hk=-k$ and $Hv = v$ for every $v\in V$).

c) Prove, by an example, that $H^2 = I$, so $H^{-1}=H$.

share|cite|improve this question

Let's talk about (a). You've got to prove that V is a sub space of $\mathbb{R}^n$, that's the idea: you can assume that the vector $k$ is $e_1=\left( \begin{array}{c} 1\\ 0\\ .\\ .\\ .\\ 0\\ \end{array} \right)$ $\space$ (why?!) and prove that $V=$Span$\{e_i, $ with $i$ from $2$ to $n$$\}$. As you can imagine its dimension is $n-1$. I suggest you to call it $k$ orthogonal or simply $k^{\perp}=V$

(b)&(c) looks a bit disconnected by (a), my interpretation it that you've got to show a matrix $H$, $n \times n$, such that $H^2=Id$ and $Hk=-k$. Consider $\begin{pmatrix}-1 & 0 \\ 0 & Id_{n-1}\end{pmatrix}$ and try to prove it, ask for any more hints.

share|cite|improve this answer
I thought we can show that something is a subspace by proving it's closed under addition and scalar multiplication? Ok, I understand that e1 can be k, because it is a standard vector. Also, what does the notation Id<sub>n-1</sub>mean? Identity matrix of n-1? How do you multiply that? – Trom Nov 24 '12 at 20:02
1) Ye, that's right, I was just following an other way (and showing the moral behind the exercise). If you want to follow the way of direct verify you won't find any difficulty and it's exactly what you said. 2) It's just a notation. It's a square $n$ matrix which has an $Id_{n-1}$ inside. – Ivan Nov 25 '12 at 7:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.