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Chapter 4: Problem 24

Consider the linear transformation \(T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}\) whose standard matrix is $$ A=\left[\begin{array}{ccc} \frac{1}{6} & \frac{1}{3}+\frac{\sqrt{6}}{6} & \frac{1}{6}-\frac{\sqrt{6}}{3} \\\ \frac{1}{3}-\frac{\sqrt{6}}{6} & \frac{2}{3} & \frac{1}{3}+\frac{\sqrt{6}}{6} \\\ \frac{1}{6}+\frac{\sqrt{6}}{3} & \frac{1}{3}-\frac{\sqrt{6}}{6} & \frac{1}{6} \end{array}\right] $$ a. Find a nonzero vector \(\mathbf{v}_{1}\) satisfying \(A \mathbf{v}_{1}=\mathbf{v}_{1}\). (Hint: Proceed as in Exercise 1.4.5.) \({ }^{7}\) b. Find an orthonormal basis \(\left\\{\mathbf{v}_{2}, \mathbf{v}_{3}\right\\}\) for the plane orthogonal to \(\mathbf{v}_{1}\). c. Let \(\mathcal{B}=\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\\}\). Apply the change-of-basis formula to find \([T]_{\mathcal{B}}\). d. Use your answer to part \(c\) to explain why \(T\) is a rotation. (Also see Example 6 in Section 1 of Chapter 7.)

Short Answer

Expert verified
In summary, we found a nonzero vector \(\mathbf{v}_1 = \left[\begin{array}{c} 1 \\ -1 \\ 1 \end{array}\right]\) that satisfies \(Av_1 = v_1\), as well as an orthonormal basis \(\mathbf{v}_2 = \frac{1}{\sqrt{2}}\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right]\) and \(\mathbf{v}_3 = \frac{1}{\sqrt{6}}\left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right]\) for the plane orthogonal to \(\mathbf{v}_1\). We applied the change of basis formula to find \([T]_{\mathcal{B}} = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ 0 & \frac{\sqrt{3}}{2} & \frac{1}{2} \end{array}\right]\), and showed that T is a rotation since ([T]B)T 脳 [T]B = I and det([T]B) = 1.

Step by step solution

01

a. Find a nonzero vector v1 satisfying Av1 = v1.

To find a nonzero vector v1 such that Av1 = v1, we must solve the following equation: \( A\mathbf{v}_1 = \mathbf{v}_1\) Using the hint, we can rewrite the equation as: \( (A - I)\mathbf{v}_1 = 0\) Where I is the identity matrix. Calculating (A - I), we get: \(A - I = \left[\begin{array}{ccc} -\frac{5}{6} & \frac{1}{3}+\frac{\sqrt{6}}{6} & \frac{1}{6}-\frac{\sqrt{6}}{3} \\ \frac{1}{3}-\frac{\sqrt{6}}{6} & -\frac{1}{3} & \frac{1}{3}+\frac{\sqrt{6}}{6} \\ \frac{1}{6}+\frac{\sqrt{6}}{3} & \frac{1}{3}-\frac{\sqrt{6}}{6} & -\frac{5}{6} \end{array}\right]\) Now, we must row reduce this matrix to find if there's a nonzero solution to v1: The row-reduced echelon form of (A - I) is: \(RREF(A - I) = \left[\begin{array}{ccc} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\) Therefore, a nonzero vector v1 that satisfies Av1 = v1 can be: \(\mathbf{v}_1 = \left[\begin{array}{c} x \\ -x \\ x \end{array}\right]\), where x is any nonzero scalar. Let x=1, then: \(\mathbf{v}_1 = \left[\begin{array}{c} 1 \\ -1 \\ 1 \end{array}\right]\)
02

b. Find an orthonormal basis {v2, v3} for the plane orthogonal to v1.

To find an orthonormal basis for the plane orthogonal to v1, we can use the Gram-Schmidt process. First, we need to find two linearly independent vectors orthogonal to v1. Let's choose one vector v2 such that it satisfies v1路v2 = 0: v2 = \(\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right]\) Now we need to find another vector v3 that is orthogonal to both v1 and v2. To find v3, we can take the cross product of v1 and v2: v3 = v1 脳 v2 = \(\left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right]\) Now, we normalize v2 and v3 to make them orthonormal: \(|\mathbf{v}_2| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{2}\) \(|\mathbf{v}_3| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}\) So the orthonormal basis {v2, v3} is: \(\mathbf{v}_2 = \frac{1}{\sqrt{2}}\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right]\ and\ \mathbf{v}_3 = \frac{1}{\sqrt{6}}\left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right]\)
03

c. Find [T]B using the change-of-basis formula.

To find [T]B (the matrix of transformation T with respect to basis B={v1, v2, v3}), we can use the change-of-basis formula: \([T]_{\mathcal{B}} = P_{\mathcal{B}}^{-1}AP_{\mathcal{B}}\) Where PB is the matrix whose columns are the basis vectors v1, v2, and v3. Therefore, PB is: \(P_{\mathcal{B}} = \left[\begin{array}{ccc} 1 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ -1 & 0 & \frac{2}{\sqrt{6}} \\ 1 & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \end{array}\right]\) We find \(P_{\mathcal{B}}^{-1}\) (the inverse of PB): \(P_{\mathcal{B}}^{-1} = \left[\begin{array}{ccc} \frac{1}{3} & -\frac{1}{3} & \frac{1}{3} \\ \frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{6}} & \frac{2}{\sqrt{6}} & \frac{1}{\sqrt{6}} \end{array}\right]\) Now we calculate [T]B: \([T]_{\mathcal{B}} = P_{\mathcal{B}}^{-1}AP_{\mathcal{B}} = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ 0 & \frac{\sqrt{3}}{2} & \frac{1}{2} \end{array}\right]\)
04

d. Show that T is a rotation.

To show that T is a rotation, we must prove that [T]B is an orthogonal matrix, which means that ([T]B)T 脳 [T]B = I when the determinant is 1. First, we find the transpose of [T]B: \(([T]_{\mathcal{B}})^T = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \frac{1}{2} & \frac{\sqrt{3}}{2} \\ 0 & -\frac{\sqrt{3}}{2} & \frac{1}{2} \end{array}\right]\) Next, we multiply ([T]B)T by [T]B: \(([T]_{\mathcal{B}})^TT_{\mathcal{B}} = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] = I\) Finally, we calculate the determinant of [T]B: \(det([T]_{\mathcal{B}}) = 1\) Since ([T]B)T 脳 [T]B = I and det([T]B) = 1, we can conclude that T is a rotation.

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Key Concepts

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

Understanding Linear Transformations
In linear algebra, a linear transformation is a mapping between two vector spaces that preserves the operations of vector addition and scalar multiplication. More technically, a transformation T is linear if for any vectors u and v in vector space V, and any scalar c, the following holds:

  • T(u + v) = T(u) + T(v)
  • T(cu) = cT(u)

The given exercise deals with such a transformation where a matrix A defines the rule for transforming vectors in 鈩澛�. Since matrix A maps 鈩澛� to itself, it's also called an endomorphism. Often, it's important to find certain vectors that are invariant under the transformation defined by matrix A, meaning the vector is mapped to a scalar multiple of itself, typically seen in eigenvalue problems.

The step-by-step solution provided begins with finding vector 饾惎鈧� which is mapped onto itself by A, implying it's an eigenvector with an eigenvalue of 1. Mathematically, this is expressed as A饾惎鈧� = 饾惎鈧�. Solving this leads us to discover the vector that remains unchanged by the linear transformation, signifying it has a special role in the vector space with respect to the transformation A.
Orthonormal Basis and Its Significance
An orthonormal basis of a vector space is a set of vectors that are both orthogonal (perpendicular) to each other and of unit length (normalized). The orthonormality condition makes computations in linear algebra particularly convenient. For instance, the orthonormal basis simplifies the process of projecting a vector onto a subspace or changing coordinates between different bases.

In our scenario, after finding 饾惎鈧�, the problem instructs us to discover an orthonormal basis for the space that is orthogonal to 饾惎鈧�. To achieve this, two additional vectors 饾惎鈧� and 饾惎鈧� are computed, ensuring that they are perpendicular to 饾惎鈧� as well as to each other. The Gram-Schmidt process is a systematic method used for orthogonalizing a set of vectors in a vector space, which was utilized here in the creation of such a basis.

The solution outlines 饾惎鈧� and 饾惎鈧� and subsequently ensures they are normalized so that they have a magnitude of one. This basis of vectors 饾惎鈧�, 饾惎鈧�, and 饾惎鈧� constitutes an orthonormal set, which greatly eases subsequent calculations and interpretations in the context of the linear transformation defined by matrix A.
Change-of-Basis Formula and Its Application
The change-of-basis formula is a pivotal concept in linear algebra that enables us to transform the representation of a linear transformation from one basis to another. A basis change is often done to simplify the problem or to obtain a representation that has specific desirable properties.

In mathematical terms, to change the basis from the standard basis to some basis 鈩�, one uses the matrix 笔冲鈩�, which is formed by placing the basis vectors as columns side by side. A transformation matrix 摆罢闭冲鈩�, representing the original linear transformation T in the new basis 鈩�, can be found by the formula:
  • 摆罢闭冲鈩� = 笔冲鈩伝鹿A笔冲鈩�

where A is the matrix representation of T in the standard basis, and 笔冲鈩伝鹿 is the inverse of 笔冲鈩�. The solution shows precisely this process by constructing 笔冲鈩� with 饾惎鈧�, 饾惎鈧�, and 饾惎鈧�, calculating its inverse, and finding 摆罢闭冲鈩�.

Knowing 摆罢闭冲鈩� is essential for understanding how the linear transformation T behaves with respect to the new basis. The resulting matrix provides deep insights into the nature of T. For instance, a special kind of transformation is where the transformed basis vectors still form an orthonormal set; such transformations include rotations and reflections, rather than dilations or shears, in the vector space. In this exercise, determining that T is a rotation involves investigating the properties of the new transformation matrix 摆罢闭冲鈩� 鈥� specifically, its orthogonality and the value of its determinant.

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Most popular questions from this chapter

Let \(V\) be the subspace of \(\mathbb{R}^{3}\) spanned by \((1,0,1)\) and \((0,1,1)\). Let \(T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}\) be the linear transformation given by reflecting across \(V\). Find the standard matrix for \(T\).

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 \(T(f)=g\) ?

Let \(V=\operatorname{Span}((1,-1,0,2),(1,0,1,1)) \subset \mathbb{R}^{4}\), and let \(\mathbf{b}=(1,-3,1,1)\). a. Find an orthogonal basis for \(V\). b. Use your answer to part \(a\) to find \(\mathbf{p}=\operatorname{proj}_{V} \mathbf{b}\). c. Letting $$ A=\left[\begin{array}{rr} 1 & 1 \\ -1 & 0 \\ 0 & 1 \\ 2 & 1 \end{array}\right] $$ use your answer to part \(b\) to give the least squares solution of \(A \mathbf{x}=\mathbf{b}\).

Suppose \(V=\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{n}\right\\}\) is an ordered basis for \(V, \mathcal{W}=\left\\{\mathbf{w}_{1}, \ldots, \mathbf{w}_{m}\right\\}\) is an ordered basis for \(W\), and \(A\) is the matrix for the linear transformation \(T: V \rightarrow W\) with respect to these bases. a. Check that \(\mathbf{x} \in \mathbf{N}(A) \Longleftrightarrow x_{1} \mathbf{v}_{1}+\cdots+x_{n} \mathbf{v}_{n} \in \operatorname{ker}(T)\). b. Check that \(\mathbf{y} \in \mathbf{C}(A) \Longleftrightarrow y_{1} \mathbf{w}_{1}+\cdots+y_{m} \mathbf{w}_{m} \in\) image \((T)\).

Let \(V\) and \(W\) be subspaces of \(\mathbb{R}^{n}\) and let \(\mathbf{b} \in \mathbb{R}^{n}\). The affine subspace passing through \(\mathbf{b}\) and parallel to \(V\) is defined to be $$ \mathbf{b}+V=\left\\{\mathbf{x} \in \mathbb{R}^{n}: \mathbf{x}=\mathbf{b}+\mathbf{v} \text { for some } \mathbf{v} \in V\right\\} . $$ a. Suppose \(\mathbf{x} \in \mathbf{b}+V\) and \(\mathbf{y} \in \mathbf{c}+W\) have the property that \(\mathbf{x}-\mathbf{y}\) is orthogonal to both \(V\) and \(W\). Show that \(\|\mathbf{x}-\mathbf{y}\| \leq\left\|\mathbf{x}^{\prime}-\mathbf{y}^{\prime}\right\|\) for any \(\mathbf{x}^{\prime} \in \mathbf{b}+V\) and \(\mathbf{y}^{\prime} \in \mathbf{c}+W\). (Thus, \(\mathbf{x}\) and \(\mathbf{y}\) are the points in \(\mathbf{b}+V\) and \(\mathbf{c}+W\), respectively, that are closest.) b. Show that the distance between the affine subspaces \(\mathbf{b}+V\) and \(\mathbf{c}+W\) (see Figure 1.3), i.e., the least possible distance between a point in one and a point in the other, is $$ \left\|\operatorname{proj}_{(V+w)^{\perp}}(\mathbf{b}-\mathbf{c})\right\| . $$ c. Deduce that this distance is 0 when \(V+W=\mathbb{R}^{n}\). Give the obvious geometric explanation.
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