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My book gives the following definition:

Let $L$ be a one dimensional subspace of $\mathbb R^2$. We may view $L$ as a line in the plane through the origin. A linear operator $T$ on $\mathbb R^2$ is called reflection of $\mathbb R^2$ about line $L$ if $T(x)=x$ for all $x \in L$ and $T(x)=-x$ for all $x \in L^\perp$.

I don't follow this definition at all, what is it trying to convey ? What is the difference between reflection about a line in $\mathbb R^2$ and reflection of $\mathbb R^2$ itself, about a line ? I am really confused.

Also what does $T(x)=x$ for all $x \in L$ and $T(x)=-x$ for all $x \in L^\perp$ mean ?

It would be nice it someone could explain this definition to me.

Also i am not very sure of what $L^\perp$ represents ? Is it just some sort of notation or does it mean the set of all those elements of $\mathbb R^2$ perpendicular to the line $L$?

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  • $\begingroup$ For the last sentence: it is both some sort of notation and means the set of all those elements of $\Bbb R^2$ perpendicular to the line $L$ (which here is supposed to pass through the origin). $\endgroup$ – Marc van Leeuwen Apr 14 '14 at 14:03
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Firslty Given a vector space $V$ and a subspace $L$, $L^{\perp}$ is the set of all vectors $l$ such that $\langle l,v \rangle=0$ for all $v \in V$.Here, $\langle , \rangle$ is a given inner prduct. Its not difficult to show that $L^{\perp}$ also has the structure of a vector space.

Now, coming to your problem, we are given a $1$-d subspace of $\mathbb{R}^2$. So, imagine it is some line $L$ in the plane passing through the origin. $T$ has to reflect every vector in the plane along this line. Its easy to visualize what exactly $T$ does. Every vector $x$ in the plane may be written uniquely as the sum of two vectors, one component along the given line $L$ and one component perpendicular to it. This is a general theorem in linear algebra and is written as $V = L \oplus L^{\perp}$.

When we reflect, the component along the line remains the same while the perpendicular component changes sign. As a simple case, consider the given line as the $x$ axis and hence its perpendicular is the $y$ axis. Now any vector in the plane may be written as the sum of its $x$ and $y$ components. After reflection, $x$ component remains same while the $y$ component changes sign. i.e. $(5,3) $becomes $(5,-3)$

The same thing is written formally as $T(x)=x$ for $x \in L$ and $T(x)=-x$ for $x \in L^{\perp}$.

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I will start with your last question, as it is paramount to understanding the rest of the definition. Given a one-dimensional subspace $L$ of $\mathbb{R}^2$, its orthogonal complement $L^{\perp}$ is the linear subspace $$L^{\perp} = \{v \in \mathbb{R}^2 \mid \forall w \in L: \langle v,w \rangle = 0 \},$$ where $\langle v, w \rangle$ denotes the standard inner product on $\mathbb{R}^2$. (This definition of the orthogonal complement generalizes to arbitrary linear subspaces of $\mathbb{R}^n$). If you don't know the inner product yet, it is perfectly fine to think of it as the unique line passing through the origin that is perpendicular to $L$.

As far as I know, all reflections of $\mathbb{R}^2$ about a line are of the form $T$. I think there is no difference in the two types of relfections you mention. Now $T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ is a linear operator on $\mathbb{R}^2$. It leaves the line $L$ exactly where it is and "flips" $L^{\perp}$ around the origin.

Let $v \in \mathbb{R}^2$ be a point. Then we can write it as a unique sum $v = v_1 + v_2$, with $v_1 \in L$ and $v_2 \in L^{\perp}$ (the lines $L$ and $L^{\perp}$ span $\mathbb{R}^2$, if that means anything to you, otherwise just think of the parallelogram law), so $T(v) = T(v_1 + v_2) = T(v_1) + T(v_2) = v_1 - v_2$. This shows that $T$ does exactly what you expect it to do, namely "flip" $\mathbb{R}^2$ around the line $L$. It is instructive to draw these things out on a piece of paper for yourself, to see what happens. Just draw two perpendicular lines, call one $L$ and the other $L^{\perp}$, and work your way from there. Maybe it helps to think of $L$ and $L^{\perp}$ as the coordinate axes of $\mathbb{R}^2$...

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