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Chapter 3: Problem 58

Let \(u_{1}=(1,1,2), u_{2}=(1,3,-2), u_{3}=(4,-2,-1)\) in \(\mathbf{R}^{3} .\) Show that \(u_{1}, u_{2}, u_{3}\) are orthogonal, and write \(v\) as a linear combination of \(u_{1}, u_{2}, u_{3},\) where (a) \(v=(5,-5,9),\) (b) \(v=(1,-3,3),\) (c) \(v=(1,1,1)\) (Hint: Use Fourier coefficients.)

Short Answer

Expert verified
The given vectors \(u_1, u_2, u_3\) are orthogonal, and for the given vectors \(v\), we have: (a) \(v=(5,-5,9) = 3u_1-2u_2+u_3\) (b) \(v=(1,-3,3) = u_1 + \frac{2}{7} u_2 + \frac{1}{3} u_3\) (c) \(v=(1,1,1) = \frac{2}{3}u_1 + \frac{1}{7}u_2 + \frac{1}{21}u_3\)

Step by step solution

01

1. Check orthogonality

To check if the given vectors are orthogonal, we will calculate the dot product of each pair of vectors and see whether they are zero or not. The dot product of \(u_1\) and \(u_2\), \(u_1\cdot u_2\): \(u_1\cdot u_2 = (1)(1) + (1)(3) + (2)(-2) = 1 + 3 - 4 = 0\) The dot product of \(u_1\) and \(u_3\), \(u_1\cdot u_3\): \(u_1\cdot u_3 = (1)(4) + (1)(-2) + (2)(-1) = 4 - 2 - 2 = 0\) The dot product of \(u_2\) and \(u_3\), \(u_2\cdot u_3\): \(u_2\cdot u_3 = (1)(4) + (3)(-2) + (-2)(-1) = 4 - 6 + 2 = 0\) Since the dot products of all pairs are zero, the given vectors are orthogonal.
02

2. Write v as the linear combination of u1, u2, u3

Now, we will use Fourier coefficients to find the linear combination of \(u_1, u_2, u_3\) that corresponds to the given vector \(v\). For any vector \(v\) in 鈩澛�, we can find the Fourier coefficients as follows: \(c_{1}=\frac{v\cdot u_{1}}{u_1\cdot u_1}\), \(c_{2}=\frac{v\cdot u_{2}}{u_2\cdot u_2}\), and \(c_{3}=\frac{v\cdot u_{3}}{u_3\cdot u_3}\). Then, the vector \(v\) can be expressed as \(v = c_1u_1 + c_2u_2 + c_3u_3\). (a) \(v = (5, -5, 9)\) Calculate the Fourier coefficients: \(c_1 = \frac{v\cdot u_1}{u_1\cdot u_1} = \frac{(5,-5,9)\cdot(1,1,2)}{(1,1,2)\cdot(1,1,2)} = \frac{5-5+18}{6}=\frac{18}{6} = 3,\) \(c_2 = \frac{v\cdot u_2}{u_2\cdot u_2} = \frac{(5,-5,9)\cdot(1,3,-2)}{(1,3,-2)\cdot(1,3,-2)} = \frac{5-15-18}{14}=\frac{-28}{14}=-2,\) \(c_3 = \frac{v\cdot u_3}{u_3\cdot u_3} = \frac{(5,-5,9)\cdot(4,-2,-1)}{(4,-2,-1)\cdot(4,-2,-1)} = \frac{20+10-9}{21} = \frac{21}{21} = 1.\) Hence, \(v=3u_1-2u_2+u_3\). (b) \(v = (1, -3, 3)\) Calculate the Fourier coefficients: \(c_1 = \frac{v\cdot u_1}{u_1\cdot u_1} = \frac{(1,-3,3)\cdot(1,1,2)}{(1,1,2)\cdot(1,1,2)} = \frac{1-3+6}{6}=1,\) \(c_2 = \frac{v\cdot u_2}{u_2\cdot u_2} = \frac{(1,-3,3)\cdot(1,3,-2)}{(1,3,-2)\cdot(1,3,-2)} = \frac{1+9-6}{14}=\frac{4}{14}= \frac{2}{7},\) \(c_3 = \frac{v\cdot u_3}{u_3\cdot u_3} = \frac{(1,-3,3)\cdot(4,-2,-1)}{(4,-2,-1)\cdot(4,-2,-1)} = \frac{4+6-3}{21} = \frac{7}{21} = \frac{1}{3}.\) Hence, \(v = u_1 + \frac{2}{7} u_2 + \frac{1}{3} u_3\). (c) \(v = (1,1,1)\) Calculate the Fourier coefficients: \(c_1 = \frac{v\cdot u_1}{u_1\cdot u_1} = \frac{(1,1,1)\cdot(1,1,2)}{(1,1,2)\cdot(1,1,2)} = \frac{1+1+2}{6}=\frac{4}{6}= \frac{2}{3},\) \(c_2 = \frac{v\cdot u_2}{u_2\cdot u_2} = \frac{(1,1,1)\cdot(1,3,-2)}{(1,3,-2)\cdot(1,3,-2)} = \frac{1+3-2}{14}=\frac{2}{14}= \frac{1}{7},\) \(c_3 = \frac{v\cdot u_3}{u_3\cdot u_3} = \frac{(1,1,1)\cdot(4,-2,-1)}{(4,-2,-1)\cdot(4,-2,-1)} = \frac{4-2-1}{21} = \frac{1}{21}.\) Hence, \(v = \frac{2}{3}u_1 + \frac{1}{7}u_2 + \frac{1}{21}u_3\).

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Key Concepts

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

Linear Algebra
In the realm of mathematics, specifically within the study of vectors and spaces, linear algebra emerges as a fundamental pillar. It deals with vectors, vector spaces, linear transformations, and systems of linear equations. Vectors are entities that have both magnitude and direction and can be added together or multiplied by scalars to produce new vectors. A vector space is a collection of vectors that can be scaled and added according to specific rules to generate new members of the space.

Within linear algebra, a key operation is the expression of a vector as a linear combination of other vectors. This means we can represent a particular vector as the sum of scaled versions of other vectors. The exercise provided is an application of this concept, where a given vector v is to be written as a combination of the vectors u1, u2, and u3.
Dot Product
The dot product, also known as scalar product, is a pivotal operation in linear algebra that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This operation is denoted by a dot between two vectors (e.g., \(\mathbf{a} \cdot \mathbf{b}\)) and is calculated by multiplying corresponding entries and then summing those products.

The significance of the dot product lies in its ability to gauge the angle between two vectors in a multi-dimensional space. If the dot product is zero, the vectors are orthogonal (perpendicular). This is exactly what was verified in step 1 of the solution: by checking whether the dot products of the given vectors were zero, we could conclude that the vectors were orthogonal to each other. In addition, the dot product is utilized in the process of finding Fourier coefficients, which are crucial for expressing a vector as a linear combination of a set of orthogonal vectors.
Fourier Coefficients
Fourier coefficients play a central role when dealing with the expression of signals, functions, or vectors in terms of a series of sinusoidal components. In the context of linear algebra, these coefficients help in representing a vector as a sum of scaled orthogonal vectors. To calculate a Fourier coefficient, one takes the dot product of the vector with an orthogonal basis vector and then normalizes it by the magnitude of the base vector squared.

In the provided solution, the Fourier coefficients enabled the expression of the vector v as a linear combination of u1, u2, and u3. By dividing the dot product of v with each ui by the dot product of ui with itself, the coefficients pertinent to each basis vector were obtained. These coefficients were then used to scale and add the orthogonal vectors to reconstruct v in the space generated by u1, u2, and u3, demonstrating the power of this mathematical tool in decomposing vectors into their components along orthogonal directions.

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Most popular questions from this chapter

Find the dimension and a basis of the general solution \(W\) of each of the following homogeneous systems: (a) \(\quad x-y+2 z=0\) \(2 x+y+z=0\) \(5 x+y+4 z=0\) (b) \(\quad x+2 y-3 z=0\) \(2 x+5 y+2 z=0\) \(3 x-y-4 z=0\) (c) \(x+2 y+3 z+t=0\) \(2 x+4 y+7 z+4 t=0\) \(3 x+6 y+10 z+5 t=0\)

Let \(e_{1}, e_{2}, e_{3}\) denote, respectively, the elementary row operations "Interchange rows \(R_{1}\) and \(R_{2}, " \quad\) "Replace \(R_{3}\) by \(7 R_{3}, " \quad\) "Replace \(R_{2}\) by \(-3 R_{1}+R_{2} "\) Find the corresponding three-square elementary matrices \(E_{1}, E_{2}, E_{3}\). Apply each operation to the \(3 \times 3\) identity matrix \(I_{3}\) to obtain $$ E_{1}=\left[\begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right], \quad E_{2}=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 7 \end{array}\right], \quad E_{3}=\left[\begin{array}{rrr} 1 & 0 & 0 \\ -3 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] $$

Consider the system $$\begin{array}{c} x+a y=4 \\ a x+9 y=b \end{array}$$ (a) For which values of \(a\) does the system have a unique solution? (b) Find those pairs of values \((a, b)\) for which the system has more than one solution.

Solve each of the following systems using its augmented matrix \(M\) : \(\begin{array}{rrr}x+2 y-z=3 & x-2 y+4 z=2 & x+y+3 z=1 \\ x+3 y+z=5 & 2 x-3 y+5 z=3 & 2 x+3 y-z=3 \\ 3 x+8 y+4 z=17 & 3 x-4 y+6 z=7 & 5 x+7 y+z=7 \\\ \text { (a) } & \text { (b) } & \text { (c) }\end{array}\) (b) (c) (a) Reduce the augmented matrix \(M\) to echelon form as follows: $$ M=\left[\begin{array}{rrrr} 1 & 2 & -1 & 3 \\ 1 & 3 & 1 & 5 \\ 3 & 8 & 4 & 17 \end{array}\right] \sim\left[\begin{array}{rrrr} 1 & 2 & -1 & 3 \\ 0 & 1 & 2 & 2 \\ 0 & 2 & 7 & 8 \end{array}\right] \sim\left[\begin{array}{rrrr} 1 & 2 & -1 & 3 \\ 0 & 1 & 2 & 2 \\ 0 & 0 & 3 & 4 \end{array}\right] $$ Now write down the corresponding triangular system $$ \begin{array}{r} x+2 y-z=3 \\ y+2 z=2 \\ 3 z=4 \end{array} $$ and solve by back-substitution to obtain the unique solution $$ x=\frac{17}{3}, y=-\frac{2}{3}, z=\frac{4}{3} \quad \text { or } \quad u=\left(\frac{17}{3},-\frac{2}{3}, \frac{4}{3}\right) $$ Alternately, reduce the echelon form of \(M\) to row canonical form, obtaining $$ M \sim\left[\begin{array}{llll} 1 & 2 & -1 & 3 \\ 0 & 1 & 2 & 2 \\ 0 & 0 & 1 & \frac{4}{3} \end{array}\right] \sim\left[\begin{array}{lllr} 1 & 2 & 0 & \frac{13}{3} \\ 0 & 1 & 0 & -\frac{2}{3} \\ 0 & 0 & 1 & \frac{4}{3} \end{array}\right] \sim\left[\begin{array}{lllr} 1 & 0 & 0 & \frac{17}{3} \\ 0 & 1 & 0 & -\frac{2}{3} \\ 0 & 0 & 1 & \frac{4}{3} \end{array}\right] $$ This also corresponds to the above solution. (b) First reduce the augmented matrix \(M\) to echelon form as follows: $$ M=\left[\begin{array}{rrrr} 1 & -2 & 4 & 2 \\ 2 & -3 & 5 & 3 \\ 3 & -4 & 6 & 7 \end{array}\right] \sim\left[\begin{array}{rrrr} 1 & -2 & 4 & 2 \\ 0 & 1 & -3 & -1 \\ 0 & 2 & -6 & 1 \end{array}\right] \sim\left[\begin{array}{rrrr} 1 & -2 & 4 & 2 \\ 0 & 1 & -3 & -1 \\ 0 & 0 & 0 & 3 \end{array}\right] $$ The third row corresponds to the degenerate equation \(0 x+0 y+0 z=3\), which has no solution. Thus, "DO NOT CONTINUE." The original system also has no solution. (Note that the echelon form indicates whether or not the system has a solution.) (c) Reduce the augmented matrix \(M\) to echelon form and then to row canonical form: $$ M=\left[\begin{array}{rrrr} 1 & 1 & 3 & 1 \\ 2 & 3 & -1 & 3 \\ 5 & 7 & 1 & 7 \end{array}\right] \sim\left[\begin{array}{rrrr} 1 & 1 & 3 & 1 \\ 0 & 1 & -7 & 1 \\ 0 & 2 & -14 & 2 \end{array}\right] \sim\left[\begin{array}{rrrr} 1 & 0 & 10 & 0 \\ 0 & 1 & -7 & 1 \end{array}\right] $$ (The third row of the second matrix is deleted, because it is a multiple of the second row and will result in a zero row.) Write down the system corresponding to the row canonical form of \(M\) and then transfer the free variables to the other side to obtain the free-variable form of the solution: $$ \begin{aligned} &x+10 z=0 \\ &y-7 z=1 \end{aligned} \quad \text { and } \quad \begin{gathered} x=-10 z \\ y=1+7 z \end{gathered} $$ Here \(z\) is the only free variable. The parametric solution, using \(z=a\), is as follows: $$ x=-10 a, y=1+7 a, z=a \quad \text { or } \quad u=(-10 a, 1+7 a, a) $$

Solve each of the following systems: \(\begin{array}{rr}x_{1}-3 x_{2}+2 x_{3}-x_{4}+2 x_{5}=2 & x_{1}+2 x_{2}-3 x_{3}+4 x_{4}=2 \\ 3 x_{1}-9 x_{2}+7 x_{3}-x_{4}+3 x_{5}=7 & 2 x_{1}+5 x_{2}-2 x_{3}+x_{4}=1 \\ 2 x_{1}-6 x_{2}+7 x_{3}+4 x_{4}-5 x_{5}=7 & 5 x_{1}+12 x_{2}-7 x_{3}+6 x_{4}=3 \\ \text { (a) } & \text { (b) }\end{array}\) Reduce each system to echelon form using Gaussian elimination: (a) Apply "Replace \(L_{2}\) by \(-3 L_{1}+L_{2}^{n}\) and "Replace \(L_{3}\) by \(-2 L_{1}+L_{3}^{n}\) to eliminate \(x\) from the second and third equations. This yields $$ \begin{aligned} x_{1}-3 x_{2}+2 x_{3}-x_{4}+2 x_{5} &=2 \\ x_{3}+2 x_{4}-3 x_{5} &=1 \\ 3 x_{3}+6 x_{4}-9 x_{5} &=3 \end{aligned} \quad \text { or } \quad x_{1}-3 x_{2}+2 x_{3}-x_{4}+2 x_{5}=2 } \\\ {x_{3}+2 x_{4}-3 x_{5}=1} \end{array} $$ (We delete \(L_{3}\), because it is a multiple of \(L_{2} .\) ) The system is in echelon form with pivot variables \(x_{1}\) and \(x_{3}\) and free variables \(x_{2}, x_{4}, x_{5}\) To find the parametric form of the general solution, set \(x_{2}=a, x_{4}=b, x_{5}=c\), where \(a, b, c\) are parameters. Back-substitution yields \(x_{3}=1-2 b+3 c\) and \(x_{1}=3 a+5 b-8 c .\) The general solution is $$ x_{1}=3 a+5 b-8 c, x_{2}=a, x_{3}=1-2 b+3 c, x_{4}=b, x_{5}=c $$ or, equivalently, \(u=(3 a+5 b-8 c, a, 1-2 b+3 c, b, c)\) (b) Eliminate \(x_{1}\) from the second and third equations by the operations "Replace \(L_{2}\) by \(-2 L_{1}+L_{2} "\) and "Replace \(L_{3}\) by \(-5 L_{1}+L_{3} . "\) This yields the system $$ \begin{array}{r} x_{1}+2 x_{2}-3 x_{3}+4 x_{4}=2 \\ x_{2}+4 x_{3}-7 x_{4}=-3 \\ 2 x_{2}+8 x_{3}-14 x_{4}=-7 \end{array} $$ The operation "Replace \(L_{3}\) by \(-2 L_{2}+L_{3} "\) yields the degenerate equation \(0=-1 .\) Thus, the system has no solution (even though the system has more unknowns than equations).
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