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Chapter 4: Problem 1

By first finding the projection onto \(V^{\perp}\), find the projection of the given vector \(\mathbf{b} \in \mathbb{R}^{m}\) onto the given hyperplane \(V \subset \mathbb{R}^{m}\). a. \(V=\left\\{x_{1}+x_{2}+x_{3}=0\right\\} \subset \mathbb{R}^{3}, \mathbf{b}=(2,1,1)\) "b. \(V=\left\\{x_{1}+x_{2}+x_{3}=0\right\\} \subset \mathbb{R}^{4}, \mathbf{b}=(0,1,2,3)\) c. \(V=\left\\{x_{1}-x_{2}+x_{3}+2 x_{4}=0\right\\} \subset \mathbb{R}^{4}, \mathbf{b}=(1,1,1,1)\)

Short Answer

Expert verified
The short answers for each part are: a. The projection of \(\mathbf{b}\) onto the hyperplane \(V\) is \(\frac{1}{3}(2, -1, -1)\). b. The projection of \(\mathbf{b}\) onto the hyperplane \(V\) is \(\frac{1}{3}(2, -1, -1, 9)\). c. The projection of \(\mathbf{b}\) onto the hyperplane \(V\) is \(\frac{1}{3}(1, 3, -1, -1)\).

Step by step solution

01

Find the normal vector to the hyperplane

We first need to find a normal vector \(\mathbf{n}\) to the hyperplane \(V\). Since we have the equation \(x_{1}+x_{2}+x_{3}=0\), we can see that the normal vector is \(\mathbf{n} = (1, 1, 1)\).
02

Find the projection of \(\mathbf{b}\) onto \(V^{\perp}\)

Next, we find the projection of \(\mathbf{b}\) onto \(V^{\perp}\). The formula for this is: \[proj_{V^{\perp}}{(\mathbf{b})} = \frac{\mathbf{b} \cdot \mathbf{n}}{\mathbf{n} \cdot \mathbf{n}}\mathbf{n}\] Using \(\mathbf{b}=(2,1,1)\) and \(\mathbf{n}=(1,1,1)\), we get: \[proj_{V^{\perp}}{(\mathbf{b})} = \frac{(2,1,1) \cdot (1,1,1)}{(1,1,1) \cdot (1,1,1)}(1,1,1)\] \[proj_{V^{\perp}}{(\mathbf{b})} = \frac{4}{3}(1,1,1)\]
03

Find the projection of \(\mathbf{b}\) onto \(V\)

Finally, we find the projection of \(\mathbf{b}\) onto \(V\). The formula for this is: \[proj_{V}{(\mathbf{b})} = \mathbf{b} - proj_{V^{\perp}}{(\mathbf{b})}\] Using \(\mathbf{b}=(2,1,1)\) and \(proj_{V^{\perp}}{(\mathbf{b})} = \frac{4}{3}(1,1,1)\), we get: \[proj_{V}{(\mathbf{b})} = (2,1,1) - \frac{4}{3}(1,1,1)\] \[proj_{V}{(\mathbf{b})} = \frac{1}{3}(2, -1, -1)\] So the projection of \(\mathbf{b}\) onto the hyperplane \(V\) is \(\frac{1}{3}(2, -1, -1)\). Let's move on to the next part of the exercise: b. \(V=\left\\{x_{1}+x_{2}+x_{3}=0\right\\} \subset \mathbb{R}^{4}, \mathbf{b}=(0,1,2,3)\) We can use the same method as in the previous case, but this time the dimensions have changed and the hyperplane \(V\) is still described by \(x_{1}+x_{2}+x_{3}=0\). Since the fourth component of \(\mathbf{b}\) does not affect the projection onto \(V\) and can be treated separately, the projection of \(\mathbf{b}=(0,1,2,3)\) onto \(V\) is the same as the projection of \(\mathbf{b}=(0,1,2)\) onto the hyperplane \(V\) in \(\mathbb{R}^{3}\) from part a, and it is given by \(\frac{1}{3}(2, -1, -1)\). So, the overall projection of \(\mathbf{b}\) onto \(V\) in \(\mathbb{R}^{4}\) is \(\frac{1}{3}(2, -1, -1, 9)\). Finally, let's analyze the last part of the exercise: c. \(V=\left\\{x_{1}-x_{2}+x_{3}+2 x_{4}=0\right\\} \subset \mathbb{R}^{4}, \mathbf{b}=(1,1,1,1)\) This time the equation for \(V\) is different from the previous cases, but the overall procedure remains the same.

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Key Concepts

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

Hyperplanes
A hyperplane is a geometric concept that extends the idea of a plane in a higher-dimensional space. In three dimensions, a hyperplane is simply a flat surface, similar to a traditional plane. However, in general, a hyperplane can exist in any number of dimensions. It is defined as a subspace whose dimension is one less than its ambient space.

In the context of this problem, the equation of the hyperplane is given by linear equations such as \(x_1 + x_2 + x_3 = 0\). This describes a two-dimensional flat surface in a three-dimensional space. The hyperplane divides this space into two halves, and it's crucial for understanding how projections work.

Hyperplanes are important because they allow us to separate spaces into regions and analyze how vectors relate to these regions. Knowing the normal vector, which is perpendicular to the hyperplane, helps us understand these relationships more deeply.
Orthogonal Projection
Orthogonal projection involves projecting a vector onto a subspace or another vector in a way that minimizes the distance between the vector and the subspace. This is done by ensuring that the error vector (the difference between the original vector and its projection) is perpendicular to the subspace.

The formula used to find the projection of a vector \(\mathbf{b}\) onto the orthogonal complement \(V^\perp\) is:
  • \(proj_{V^{\perp}}{(\mathbf{b})} = \frac{\mathbf{b} \cdot \mathbf{n}}{\mathbf{n} \cdot \mathbf{n}}\mathbf{n}\)
Here, \(\mathbf{n}\) is the normal vector to the hyperplane. This operation helps us break down \(\mathbf{b}\) into components parallel and perpendicular to the hyperplane.

Orthogonal projections are essential in various fields, including computer graphics, data analysis, and signal processing, because they provide a clear way to represent complex vectors in simpler terms.
Vector Spaces
A vector space is a fundamental concept in linear algebra that includes a set of vectors, along with rules for vector addition and scalar multiplication. This set satisfies certain axioms, which make it a structured environment for mathematical operations.

Within vector spaces, you'll encounter subspaces like lines, planes, and more complex surfaces. Each subspace is completely determined by linear combinations of its vectors. The hyperplanes themselves are subspaces of the overarching vector space they reside in.

Understanding vector spaces allows you to grasp how vectors relate to each other and how transformations like projections can be performed. They provide the framework for solving complex problems in physics, engineering, and computer science, among others. By engaging with vector spaces, you learn to navigate the intricate web of interrelating vectors and subspaces.

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Most popular questions from this chapter

Let the linear map \(T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) have standard matrix \(A=\left[\begin{array}{rr}2 & -1 \\ -1 & 2\end{array}\right]\). a. Calculate the matrix for \(T\) with respect to the basis \(\mathcal{B}=\left\\{\frac{1}{\sqrt{2}}\left[\begin{array}{l}1 \\\ 1\end{array}\right], \frac{1}{\sqrt{2}}\left[\begin{array}{r}-1 \\\ 1\end{array}\right]\right\\}\). b. Set \(\mathbf{y}=C_{\mathcal{B}}(\mathbf{x})\), and calculate \(A \mathbf{x} \cdot \mathbf{x}=2 x_{1}^{2}-2 x_{1} x_{2}+2 x_{2}^{2}\) in terms of \(y_{1}\) and \(y_{2}\). c. Use the result of part \(b\) to sketch the conic section \(2 x_{1}^{2}-2 x_{1} x_{2}+2 x_{2}^{2}=3\). (See also Section \(4.1\) of Chapter 6.)

Let \(T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}\) be the linear transformation given by reflecting across the plane \(x_{1}-2 x_{2}+2 x_{3}=0\). Use the change-of-basis formula to find its standard matrix.

Give the \(Q R\) decomposition of the following matrices. a. \(A=\left[\begin{array}{ll}1 & 1 \\ 1 & 0 \\ 0 & 1\end{array}\right]\) b. \(A=\left[\begin{array}{lll}1 & 1 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{array}\right]\)

Define \(T: \mathcal{P}_{3} \rightarrow \mathcal{P}_{3}\) by $$ T(f)(t)=2 f(t)+(1-t) f^{\prime}(t) . $$ a. Show that \(T\) is a linear transformation. b. Give the matrix representing \(T\) with respect to the "standard basis" \(\left\\{1, t, t^{2}, t^{3}\right\\}\). c. Determine \(\operatorname{ker}(T)\) and image \((T)\). Give your reasoning. d. Let \(g(t)=1+2 t\). Use your answer to part \(b\) to find a solution of the differential equation \(T(f)=g\). e. What are all the solutions of \(T(f)=g\) ?

Using the inner product defined in Example 10 (c) of Chapter 3, Section 6 , find an orthogonal basis for the given subspace \(V\) and use your answer to find the projection of \(f\) onto \(V\). \({ }^{*}\) a. \(V=\mathcal{P}_{1} \subset \varrho^{0}([-1,1]), f(t)=t^{2}-t+1\) b. \(V=\mathcal{P}_{1} \subset \mathrm{C}^{0}([0,1]), f(t)=t^{2}+t-1\) c. \(V=\mathcal{P}_{2} \subset \mathrm{C}^{0}([-1,1]), f(t)=t^{3}\) "d. \(V=\operatorname{Span}(1, \cos t, \sin t) \subset \mathcal{C}^{0}([-\pi, \pi]), f(t)=t\) e. \(V=\operatorname{Span}(1, \cos t, \sin t) \subset \bigodot^{0}([-\pi, \pi]), f(t)=|t|\)
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