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Chapter 2: Problem 13

Let \(\ell\) be a line through the origin in \(\mathbb{R}^{2}\). a. Show that \(P_{\ell}^{2}=P_{\ell} \circ P_{\ell}=P_{\ell}\). b. Show that \(R_{\ell}^{2}=R_{\ell} \circ R_{\ell}=I\).

Short Answer

Expert verified
To summarize the solution: a. For any point \(x \in \mathbb{R}^2\), we showed that projecting \(x\) onto line \(\ell\) twice gives the same result as projecting \(x\) onto line \(\ell\) once. Thus, \(P_{\ell}^2=P_{\ell}\circ P_{\ell}=P_{\ell}\). b. For any point \(x \in \mathbb{R}^2\), we showed that reflecting \(x\) across line \(\ell\) twice gives the identity transformation, taking \(x\) back to itself. Thus, \(R_{\ell}^2=R_{\ell}\circ R_{\ell}=I\).

Step by step solution

01

Let any point be \(x\)

Choose any point \(x \in \mathbb{R}^2\).
02

Compute the projection of \(x\) onto \(\ell\)

We have to project \(x\) onto the line \(\ell\), which gives us \(P_{\ell}(x)\). Let this point be \(p\).
03

Compute the projection of \(p\) onto \(\ell\)

Notice that \(p\) is already on the line \(\ell\) since we projected \(x\) onto the line. Therefore, if we project it again onto \(\ell\), it will remain unchanged since \(p\) already lies on \(\ell\). That is, \(P_{\ell}(p)=p\).
04

Conclude the result

We've shown that if we first project \(x\) onto line \(\ell\) to get \(p\) and then project \(p\) again onto the line \(\ell\), we get the same result, \(p\). Thus, \(P_{\ell}^2=P_{\ell}\circ P_{\ell}=P_{\ell}\). b. Proving \(R_{\ell}^2=R_{\ell} \circ R_{\ell}=I\).
05

Compute the reflection of \(x\) across \(\ell\)

Reflect point \(x\) across the line \(\ell\), giving us the point \(R_{\ell}(x)\). Let this point be \(r\).
06

Compute the reflection of \(r\) across \(\ell\)

Reflect point \(r\) across the line \(\ell\), which we will denote as \(R_{\ell}(r)\).
07

Prove that \(R_{\ell}(r)=x\)

Since both \(x\) and \(r\) are reflections of each other across the line \(\ell\), reflecting \(r\) across the line \(\ell\) will give us back the original point \(x\). Therefore, \(R_{\ell}(r)=x\).
08

Conclude the result

We've shown that if we first reflect \(x\) across the line \(\ell\) to get \(r\), and then reflect \(r\) back across the line \(\ell\), we obtain the original point \(x\). Thus, \(R_{\ell}^2=R_{\ell}\circ R_{\ell}=I\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Projection Onto a Line
Imagine you're standing under the sun with a stick; the shadow it casts on the ground is akin to the concept of a projection. In linear algebra, when we talk about the projection of a vector onto a line, we are referring to the 'shadow' that vector casts onto the line when we 'shine a light' perpendicularly from the vector down to the line.

Formally, if you have a vector and a line through the origin, the projection of the vector onto the line is the point where a line drawn through the vector at a right angle to the original line intersects it. This projection captures only the component of the vector that lies along the line. As per the step-by-step solution given, projecting a point onto the line twice is redundant since the point after the first projection will not move鈥攊t's already on the line, just like the stick's shadow doesn't become 'more shadowy' if you kept focusing light on it.
Reflection Across a Line
Envision a quiet water body reflecting the landscape around it; this nature's act is what we mathematically describe as a reflection. In linear algebra, reflecting a vector across a line is like flipping it over that line.

The spot where it lands is a mirror image with respect to the line, equidistant but on the opposite side. Following our exercise, reflecting a point across a line and then reflecting once more will bring the point back to its original spot, which is why the result of performing the reflection operation twice equals the identity matrix (it preserves the original vector, just as you'd look the same if you looked into a mirror that reflects a reflection of you).
Linear Transformations
Walking through a hall of mirrors at a funhouse shows how an image can be stretched, shrunk, or flipped, demonstrating effects that can be likened to linear transformations. These transformations are functions between vector spaces that preserve vector addition and scalar multiplication鈥攖hink of them as systematic ways to manipulate vectors following certain rules.

Both projections and reflections are examples of linear transformations. They alter vectors in a linear fashion: projections squash them onto a line, and reflections flip them across a line. These operations can be represented with matrices that, when applied to vectors, produce the transformed vectors. Our exercise shows these concepts at work: the projection and reflection operations applied to a vector produce predictable results consistent with their respective definitions.
Identity Matrix
The identity matrix in linear algebra is like a photographic negative in an old camera - it perfectly preserves the image with no distortions or color changes. Analogously, the identity matrix is the mathematical equivalent of 'do nothing to the vector'.

It is represented by a square matrix filled with zeros, except for 1s on the diagonal from the upper left corner to the lower right corner. Symbolically, we denote it as 'I'. In terms of linear transformations, when you apply the identity matrix to a vector, it leaves it unchanged. As demonstrated in the reflection example from our exercise, applying a reflection twice is akin to applying the identity matrix, returning the vector to its original state.

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Most popular questions from this chapter

(Recall the definition of orthogonal matrices from Exercise 19.) a. Show that if \(A\) and \(B\) are orthogonal \(n \times n\) matrices, then so is \(A B\). *b. Show that if \(A\) is an orthogonal matrix, then so is \(A^{-1}\).

a. Show that the matrix \(\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\) has no \(L U\) decomposition. b. Show that for any \(m \times n\) matrix \(A\), there is an \(m \times m\) permutation matrix \(P\) so that \(P A\) does have an \(L U\) decomposition.

Let \(A\) be an \(m \times n\) matrix. a. Assume the rank of \(A\) is \(m\) and \(B\) is a right inverse of \(A\). Show that \(B^{\prime}\) is another right inverse of \(A\) if and only if \(A\left(B-B^{\prime}\right)=\mathrm{O}\) and that this occurs if and only if every column of \(B-B^{\prime}\) is orthogonal to every row of \(A\). b. Assume the rank of \(A\) is \(n\) and \(C\) is a left inverse of \(A\). Show that \(C^{\prime}\) is another left inverse of \(A\) if and only if \(\left(C-C^{\prime}\right) A=\mathrm{O}\) and that this occurs if and only if every row of \(C-C^{\prime}\) is orthogonal to every column of \(A\).

Suppose that \(T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{2}\) is a linear transformation and that \(T\left(\left[\begin{array}{l}1 \\ 2 \\\ 1\end{array}\right]\right)=\left[\begin{array}{r}-1 \\ 2\end{array}\right]\) and \(T\left(\left[\begin{array}{r}2 \\ -1 \\\ 1\end{array}\right]\right)=\left[\begin{array}{l}3 \\ 0\end{array}\right]\) Compute \(T\left(2\left[\begin{array}{r}2 \\ -1 \\\ 1\end{array}\right]\right), T\left(\left[\begin{array}{l}3 \\ 6 \\\ 3\end{array}\right]\right)\), and \(T\left(\left[\begin{array}{r}-1 \\ 3 \\\ 0\end{array}\right]\right)\).

a. Suppose \(A \in \mathcal{M}_{m \times n}, B \in \mathcal{M}_{n \times m}\), and \(B A=I_{n}\). Prove that if for some \(\mathbf{b} \in \mathbb{R}^{m}\) the equation \(A \mathbf{x}=\mathbf{b}\) has a solution, then that solution is unique. b. Suppose \(A \in \mathcal{M}_{m \times n}, C \in \mathcal{M}_{n \times m}\), and \(A C=I_{m}\). Prove that the system \(A \mathbf{x}=\mathbf{b}\) is consistent for every \(\mathbf{b} \in \mathbb{R}^{m}\). $$ \begin{aligned} (A+B)^{k} &=\sum_{i=0}^{k} \frac{k !}{i !(k-i) !} A^{k-i} B^{i} \\ &=A^{k}+k A^{k-1} B+\frac{k(k-1)}{2} A^{k-2} B^{2}+\frac{k(k-1)(k-2)}{6} A^{k-3} B^{3} \\ &+\cdots+k A B^{k-1}+B^{k} \end{aligned} $$ Show that the result is false when \(A B \neq B A\).
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