Prove reflection in a hyperplane is a linear map

Let $\alpha \in \mathbb{R}^n$, $n \geq 2$, be a non-zero vector. Define a reflection in the hyperplane perpendicular to $\alpha$ by: $$\sigma_{\alpha}(v) = v - \dfrac{2(v, \alpha)}{(\alpha, \alpha)} \cdot \alpha$$ ($(x, y)$ is the usual inner product on $\mathbb{R}^n$).

1) Show $\sigma_{\alpha}$ is a linear map that fixes the hyperplane orthogonal to $\alpha$ and sends $\alpha$ to $-\alpha$.

2) Given $\alpha, \beta$ non-zero vectors, determine when the subgroup $\langle \sigma_{\alpha}, \sigma_{\beta} \rangle$ is infinite. Find its order when it is finite.

For 2) I don't understand what the group is. If $\sigma_{\alpha}$ and $\sigma_{\beta}$ are elements of a group, what other elements do they generate? Like for example, $\sigma_{\alpha}(\sigma_{\beta}(v)) = \left(v - \dfrac{2(v, \beta)}{(\beta, \beta)} \cdot \beta \right) - \dfrac{2\left(v - \dfrac{2(v, \beta)}{(\beta, \beta)} \cdot \beta, \beta \right)}{(\beta, \beta)} \cdot \beta$ which I guess makes sense (in the sense that dot products work in this function since the dot product is between vectors). But how do I know when there will be an infinite many number of these, and when there will be finitely many?

I can't even find an identity function $\sigma$, because a composition of $\sigma_{\alpha}$ and $\sigma_{\beta}$ is $\sigma_{\beta}$ only when $\sigma_{\alpha} = v$, but this is a constant function and does not reflect $\alpha$ about the hyperplane to $-\alpha$, so this constant function cannot be in the group.

• I don't want to keep pinging Lee by using his nice answer. Essentially, the span of $\alpha$ and $\beta$ should tell you everything you need to know. Just think about two vectors in $\Bbb R^2$, and think about reflecting across their normals. Keep dihedral groups in mind. If I can do it well, I may elaborate in an answer (probably not soon). – pjs36 Sep 23 '15 at 22:06
• @pjs36 If the two vectors are not normal to each other, then if you reflect a vector $\alpha$ across its normal, this reflected vector is not normal to the other vector $\beta$ so you can't reflect across the normal of $\beta$, right? – mr eyeglasses Sep 23 '15 at 22:30

Get some intuition from three dimensions first. Say the intersection of the two planes is the axis spanned by $\gamma$. Then $\{\alpha,\beta,\gamma\}$ is a basis, and the reflections only act on the $\alpha$ and $\beta$ components of any vector. This generalizes: prove that ${\rm span}\{\alpha,\beta\}$ is the orthogonal complement of the planes' intersection.

(More generally, $A^\perp\cap B^\perp=(A+B)^\perp$ for any subspaces $A,B$ of an inner product space.)

So really, you only need to worry about what the reflections do to the plane $\alpha$ and $\beta$ generate. That's only two dimensions to worry about. Without loss of generality, say one reflection is across the $x$-axis and the other is across the line $y=\tan(\theta)x$ (which makes an angle of $\theta$ with the $x$-axis). What exactly is the composition of the two reflections then?

If it helps, draw these two lines on a piece of paper, and put a point $P$ just under the $x$-axis in the fourth quadrant. Reflect across the $x$-axis to get a point $Q$, then reflect across the other line to get point $R$. If you label all of the angles made (between the lines and the imaginary line segments joining the origin to the three points) you should be able to make some deductions about the angles, and then get an idea for what the composition of the two reflections is.

(Spoiler: you'll be thinking about $n$-gons and dihedral groups soon after that.)

• Say one hyperplane is on the $x-axis$ and another hyperplane is $\tan(\theta)x$. Then if a vector is on the $y-axis$, then it gets reflected to the other side of the $y-axis$ by the hyperplane on the $x-axis$ because it is orthogonal to it. Then this vector doesn't get reflected across the $\tan(\theta)x$ hyperplane because it isn't orthogonal to it. So the composition of the two reflections is just one of the reflections since the other reflection doesn't do anything to it. But then I get a contradiction in that every element of the group is the identity. – mr eyeglasses Sep 25 '15 at 11:28
• For example, look at my picture: i.imgur.com/j3Ns9kL.png The red and blue lines are the hyperplane, and the orange is a vector. The orange vector gets reflected across the red hyperplane because it is orthogonal to the red hyperplane, but it does not get reflected across the blue hyperplane because it is not orthogonal to it. So the composition doesn't do anything – mr eyeglasses Sep 25 '15 at 11:50
• @morphic We have a big problem: you don't know what a reflection is. When applying a reflection across a plane, the only points in all of space that don't get moved are the points on that plane. Reflections flip all other points to their mirror images across the plane - the nature of this operation should be geometrically obvious. Based on my reading of the comments under Lee's answer, you're not simply confused about where to go next, you're also saying a lot of things that don't make any sense or are just plain wrong - I suggest reviewing all the basic concepts that are in play here. – whacka Sep 25 '15 at 13:30
• I understand a general reflection across a plane, but in the problem it states that "...reflection in the hyperplane $\textbf{perpendicular to}$..." so I thought that meant it reflects vectors across the hyperplane only when the vector is perpendicular to the hyperplane. I guess I am misunderstanding? – mr eyeglasses Sep 25 '15 at 13:34
• The hyperplane itself is perpendicular to the given vector. It's not saying the only vectors that get moved are the ones perpendicular to the vectors that make up the plane. That wouldn't be a reflection (or even a continuous map, since it permutes points on a line but fixes every point on a tubular neighborhood around the line). If $\alpha$ is a vector, then $\alpha^\perp$ is a plane (through the origin), and by definition $\sigma_\alpha$ is reflection across $\alpha^\perp$. Also, as a good exercise, you should verify the formula it gives for $\sigma_\alpha$ if at all possible. – whacka Sep 25 '15 at 13:36

For your first subquestion of 1), the hyperplane is described in the question: it is the hyperplane orthogonal to $\alpha$. You know from linear algebra that this hyperplane is the solution of the equation $\alpha \cdot v = 0$. So your goal is to take any $v$ in that hyperplane, i.e. take any $\nu$ such that $\alpha\cdot v=0$, and prove the equation $\sigma_\alpha(v)=v$.

For your second subquestion of 1), $\alpha$ is a vector, and it is a constant, so it is a constant vector (unlike $v$ which is a vector, and it is a variable, so it is a variable vector). To say that a function $f$ sends $a$ to $b$ means $f(a)=b$. So to say that the function $\sigma_\alpha$ sends $\alpha$ to $-\alpha$ means that $\sigma_\alpha(\alpha)=-\alpha$. That's the equation you are asked to prove.

For 2), you are correct that a function is not a group, but the question does not ask you to believe that a function is a group. Instead, the question asks you to believe that the set of all linear isomorphisms of $\mathbb{R}^n$ is a group under the binary operation of composition --- you may have heard of this group, it is denoted $GL(n,\mathbb{R})$. Also, you are asked to believe that if $\alpha$ is a constant vector then $\sigma_\alpha$ is an element of the group $GL(n,\mathbb{R})$. Also, if you fix two constant vectors $\alpha$ and $\beta$, then you are asked to believe that there is a subgroup of $GL(n,\mathbb{R})$ denoted $\langle \sigma_\alpha,\sigma_\beta \rangle$ and called the subgroup of $GL(n,\mathbb{R})$ that is generated by $\sigma_\alpha,\sigma_\beta$.

• Thank you. I'm only having trouble visualizing the subgroup now. I don't understand what $\sigma_\alpha,\sigma_\beta$ generates. Or even what $\sigma_\alpha$ generates. I know that $GL(n, \mathbb{R})$ is the group of invertible matrices of size $n x n$. I actually don't have much linear algebra background, but I believe these matrices can represent linear maps. Anyway, I can't seem to visualize functions like $\sigma_\alpha$ as matrices. So if $\alpha$ is some fixed vector, then – mr eyeglasses Sep 23 '15 at 20:37
• then $\sigma_\alpha$ reflects the hyperplane orthogonal to $\alpha$ (sort of like flipping the hyperplane, but retaining its position). So if we have two of these fixed vectors $\alpha$ and $\beta$, we have two different hyperplanes, unless $\alpha$ and $\beta$ are parallel to each other in which case we have the same hyperplane. What exactly are the group elements here? $\sigma_\alpha(v)$ for each $v$ is an element of the group? – mr eyeglasses Sep 23 '15 at 20:39
• No, @morphic, $\sigma_\alpha(v)$ is a vector, your group elements should be linear maps; things that move vectors around. I'll say that I think you should focus on the subspace $\operatorname{span}(\alpha, \beta)$, as it will be fixed by the group generated by those reflections. – pjs36 Sep 23 '15 at 21:43
• @pjs36 Sorry I meant that $\sigma_\alpha$ is an element (since it is a linear map). What is the intuition behind when the subgroup has infinite or finite order? – mr eyeglasses Sep 23 '15 at 21:55
• Look at the formula for $\sigma_\alpha$. Then look at the formula for $\sigma_\alpha^2 = \sigma_\alpha \circ \sigma_\alpha$. Then look at the formula for $\sigma_\alpha^3 = \sigma_\alpha^2 \circ \sigma_\alpha$. Then look at ………………………. And, an infinite amount of time later when you've got all those formulas in front of you, compare the results to the formula for the identity element of the group. – Lee Mosher Sep 23 '15 at 22:37