/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Let \(S\) be the set of all \(\l... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 18

Let \(S\) be the set of all \(\left(t_{1}, t_{2}, t_{3}\right) \in \mathrm{R}^{3}\) for which $$ 3 t_{1}^{2}+3 t_{2}^{2}+3 t_{3}^{2}-2 t_{1} t_{3}+2 \sqrt{2}\left(t_{1}+t_{3}\right)+1=0 $$ Find an orthonormal basis \(\beta\) for \(\mathrm{R}^{3}\) for which the equation relating the coordinates of points of \(\mathcal{S}\) relative to \(\beta\) is simpler. Describe \(\mathcal{S}\) geometrically.

Short Answer

Expert verified
The orthonormal basis \(\beta\) for \(\mathrm{R}^{3}\) is given by \(\left\{\frac{1}{\sqrt{2}}(1, 0, 1), \frac{1}{\sqrt{2}}(1, 0, -1), (0, 1, 0)\right\}\). The geometric shape of \(\mathcal{S}\) is a sphere with radius 1 centered at the origin.

Step by step solution

01

Rewrite the equation in standard form

First, we need to write the given equation in standard form. We will group the terms and factor the constants accordingly. The equation $$ 3 t_{1}^{2}+3 t_{2}^{2}+3 t_{3}^{2}-2 t_{1} t_{3}+2 \sqrt{2}\left(t_{1}+t_{3}\right)+1=0 $$ can be rewritten as $$ 3(t_{1}^2 + t_{2}^2 + t_{3}^2) - 2t_{1}t_{3} +2\sqrt{2}(t_{1} + t_{3}) + 1 = 0 $$
02

Find an orthonormal basis

Now, we will find an orthonormal basis 尾 for R^3 for which the equation relating the coordinates of points of S is simpler. To do this, we will represent the given equation as a quadratic form and then find the eigenvectors and eigenvalues to construct the orthonormal basis. Let $$ A = \begin{bmatrix} 3 & 0 & -1 \\ 0 & 3 & 0 \\ -1 & 0 & 3 \end{bmatrix} $$ be the matrix representing the given quadratic form. We will now find the eigenvalues 位 and eigenvectors v of matrix A. Compute the characteristic polynomial of A, $$ \det(A - \lambda I) = \begin{vmatrix} 3-\lambda & 0 & -1 \\ 0 & 3-\lambda & 0 \\ -1 & 0 & 3-\lambda \end{vmatrix}=(3-\lambda)^3 - 2(3-\lambda) = (\lambda-1)(\lambda-4)^2 $$ The eigenvalues are 1 and 4. For 位 = 1, find the eigenvectors by solving (A - 位I)v = 0: $$ \begin{bmatrix} 2 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 2 \end{bmatrix}v = 0 $$ We have the eigenvector v鈧� = (1, 0, 1). For 位 = 4, find the eigenvectors by solving (A - 位I)v = 0: $$ \begin{bmatrix} -1 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & -1 \end{bmatrix}v = 0 $$ We have two eigenvectors, v鈧� = (1, 0, -1) and v鈧� = (0, 1, 0). Now, we will construct an orthonormal basis 尾 using the eigenvectors as basis vectors: $$ \beta = \left\{\frac{v鈧亇{\|v鈧乗|}, \frac{v鈧倉{\|v鈧俓|}, \frac{v鈧儅{\|v鈧僜|}\right\}= \left\{\frac{1}{\sqrt{2}}(1, 0, 1), \frac{1}{\sqrt{2}}(1, 0, -1), (0, 1, 0)\right\} $$
03

Describe the geometric shape

With the basis 尾, the equation relating the coordinates of points of S can be rewritten as: $$ (t_{1}^{2} + t_{2}^{2} + t_{3}^{2}) - 1 = 0 $$ The geometric shape of S can be identified as a sphere with radius 1 centered at the origin.

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Key Concepts

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

Quadratic Forms
A quadratic form in linear algebra is an expression involving the sum of squared variables, each multiplied by a constant coefficient. In the context of our problem, the quadratic form is written as: \[ \quad 3t_1^2 + 3t_2^2 + 3t_3^2 - 2t_1t_3 + 2\sqrt{2}(t_1 + t_3) + 1 = 0 \] These forms are particularly useful in describing geometric shapes and can be expressed in a matrix. Here, the matrix \( A \) represents the quadratic form: \[ \quad A = \begin{bmatrix} 3 & 0 & -1 \ 0 & 3 & 0 \ -1 & 0 & 3 \end{bmatrix} \] Quadratic forms help us understand more complex equations by simplifying them, such as turning our original equation into a simpler one describing a sphere through further transformation.
Eigenvectors
Eigenvectors are special vectors associated with a matrix. When transformed by the matrix, they only scale in magnitude but do not change direction. This transformation property makes them central to solving quadratic forms. For our matrix \( A \), we calculate the eigenvectors by solving the equation: \[ \quad (A - \lambda I)v = 0 \] where \( \lambda \) are the eigenvalues, and \( v \) are the eigenvectors. Through this process, we found:\[ \quad v_1 = (1, 0, 1), \quad v_2 = (1, 0, -1), \quad v_3 = (0, 1, 0) \] These eigenvectors are the cornerstone for establishing a new coordinate system that simplifies our quadratic form's geometric representation.
Orthonormal Basis
An orthonormal basis in linear algebra is a set of vectors that are both orthogonal and normalized. Orthogonal means each pair of vectors is perpendicular, and normalized means each vector has a unit length. Using the eigenvectors \( v_1, v_2, \) and \( v_3 \) from our matrix:
  • We normalize them to ensure they have a unit length.
  • This involves dividing each vector by its magnitude.
Thus, our orthonormal basis \( \beta \) becomes: \[ \quad \beta = \left\{ \frac{1}{\sqrt{2}}(1, 0, 1), \frac{1}{\sqrt{2}}(1, 0, -1), (0, 1, 0) \right\} \] This basis transforms our original set \( S \) into a simpler form, making it easier to analyze geometrically.
Eigenvalues
Eigenvalues are scalar values in linear algebra that indicate how much their corresponding eigenvectors are stretched or shrunk during a transformation. They are critical when simplifying quadratic forms through matrices like \( A \). The process to find these values involves solving the characteristic polynomial, set by the determinant: \[ \quad \det(A - \lambda I) = 0 \] For our problem, the eigenvalues are:
  • \( \lambda_1 = 1 \)
  • \( \lambda_2 = 4 \)
These eigenvalues help identify our orthonormal basis, which allows the original complex space to transform into a geometrically simple shape like a sphere. Eigenvalues essentially unlock the hidden simplicity within the quadratic form.

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Most popular questions from this chapter

The following definition is used in Exercises 26-30. Definition. Let \(\mathrm{V}\) be a vector space over \(F\), where \(F\) is either \(R\) or \(C\). Regardless of whether \(\mathrm{V}\) is or is not an inner product space, we may still define a norm \(\|\cdot\|_{V}\) as a real-valued function on \(\mathrm{V}\) satisfying the following three conditions for all \(x, y \in \mathrm{V}\) and \(a \in F\) : (1) \(\|x\|_{v} \geq 0\), and \(\|x\|_{v}=0\) if and only if \(x=0\). (2) \(\|a x\|_{V}=|a| \cdot\|x\|_{V}\). (3) \(\|x+y\|_{v} \leq\|x\|_{v}+\|y\|_{v}\). Let \(\|\cdot\|_{V}\) be a norm (as defined on page 337 ) on a complex vector space V satisfying the parallelogram law given in Exercise 11 . Prove that there is an inner product \(\langle\cdot, \cdot\rangle\) on \(\mathrm{V}\) such that \(\|x\|_{v}^{2}=\langle x, x\rangle\) for all \(x \in \mathrm{V}\). Hint: Apply Exercise 29 to \(\mathrm{V}\) regarded as a vector space over \(R\). Then apply Exercise 25 .

Prove the Cayley-Hamilton theorem for a complex \(n \times n\) matrix \(A\). That is, if \(f(t)\) is the characteristic polynomial of \(A\), prove that \(f(A)=O\). Hint: Use Schur's theorem to show that \(A\) may be assumed to be upper triangular, in which case $$ f(t)=\prod_{i=1}^{n}\left(A_{i i}-t\right) . $$ Now if \(\mathrm{T}=\mathrm{L}_{A}\), we have \(\left(A_{j j} \mid-\mathrm{T}\right)\left(e_{j}\right) \in \operatorname{span}\left(\left\\{e_{1}, e_{2}, \ldots, e_{j-1}\right\\}\right)\) for \(j \geq 2\), where \(\left\\{e_{1}, e_{2}, \ldots, e_{n}\right\\}\) is the standard ordered basis for \(\mathrm{C}^{n}\). (The general case is proved in Section 5.4.)

Determine which of the mappings that follow are bilinear forms. Justify your answers. (a) Let \(\mathrm{V}=\mathrm{C}[0,1]\) be the space of continuous real-valued functions on the closed interval \([0,1]\). For \(f, g \in \mathrm{V}\), define $$ H(f, g)=\int_{0}^{1} f(t) g(t) d t . $$ (b) Let \(\mathrm{V}\) be a vector space over \(F\), and let \(J \in \mathcal{B}(\mathrm{V})\) be nonzero. Define \(H: \mathrm{V} \times \mathrm{V} \rightarrow F\) by $$ H(x, y)=[J(x, y)]^{2} \quad \text { for all } x, y \in \mathrm{V} . $$ (c) Define \(H: R \times R \rightarrow R\) by \(H\left(t_{1}, t_{2}\right)=t_{1}+2 t_{2}\). (d) Consider the vectors of \(\mathrm{R}^{2}\) as column vectors, and let \(H: \mathrm{R}^{2} \rightarrow R\) be the function defined by \(H(x, y)=\operatorname{det}(x, y)\), the determinant of the \(2 \times 2\) matrix with columns \(x\) and \(y\). (e) Let \(\mathrm{V}\) be a real inner product space, and let \(H: \mathrm{V} \times \mathrm{V} \rightarrow R\) be the function defined by \(H(x, y)=\langle x, y\rangle\) for \(x, y \in \mathrm{V}\). (f) Let \(\mathrm{V}\) be a complex inner product space, and let \(H: \mathrm{V} \times \mathrm{V} \rightarrow C\) be the function defined by \(H(x, y)=\langle x, y\rangle\) for \(x, y \in \mathrm{V}\).

For each of the following matrices \(A\) with entries from \(R\), find a diagonal matrix \(D\) and an invertible matrix \(Q\) such that \(Q^{t} A Q=D\). (a) \(\left(\begin{array}{ll}1 & 3 \\ 3 & 2\end{array}\right)\) (b) \(\left(\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right)\) (c) \(\left(\begin{array}{rrr}3 & 1 & 2 \\ 1 & 4 & 0 \\ 2 & 0 & -1\end{array}\right)\) Hint for \((b)\) : Use an elementary operation other than interchanging columns.

Let \(V\) be a complex inner product space, and let \(T\) be a linear operator on \(\mathrm{V}\). Define $$ \mathrm{T}_{1}=\frac{1}{2}\left(\mathrm{~T}+\mathrm{T}^{*}\right) \quad \text { and } \quad \mathrm{T}_{2}=\frac{1}{2 i}\left(\mathrm{~T}-\mathrm{T}^{*}\right) . $$ (a) Prove that \(\mathrm{T}_{1}\) and \(\mathrm{T}_{2}\) are self-adjoint and that \(\mathrm{T}=\mathrm{T}_{1}+i \mathrm{~T}_{2}\). (b) Suppose also that \(\mathrm{T}=\mathrm{U}_{1}+i \mathrm{U}_{2}\), where \(\mathrm{U}_{1}\) and \(\mathrm{U}_{2}\) are self-adjoint. Prove that \(\mathrm{U}_{1}=\mathrm{T}_{1}\) and \(\mathrm{U}_{2}=\mathrm{T}_{2}\). (c) Prove that \(T\) is normal if and only if \(T_{1} T_{2}=T_{2} T_{1}\).
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