Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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|improve this question
    
I think you will probably be downvoted soon unless you change some things: first, your question is not very well stated. "We define a matrix, where..". So, what is the matrix that we defined? Also you are using "n" as a vector of $\mathbb{R}^n$. It's better not to use the same letter for different things (although it's boldface). Another thing is that if you plan to continue using this site, invest a little bit of time learning the latex syntax. You may also want to use the "homework" tag here.. –  geo909 Nov 24 '12 at 19:06
    
Thank you for your remarks. I hope I made the text a little more clearer now. –  Trom Nov 24 '12 at 19:24
add comment

1 Answer

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|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
add comment

Your Answer

 
discard

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.