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Let \(\left\\{\mathbf{e}_{1}, \ldots, \mathbf{e}_{n}\right\\}\) be an orthogonal basis of \(\mathbb{R}^{n} .\) Given \(\mathbf{x}\) and \(\mathbf{y}\) in \(\mathbb{R}^{n},\) show that $$ \mathbf{x} \cdot \mathbf{y}=\frac{\left(\mathbf{x} \cdot \mathbf{e}_{1}\right)\left(\mathbf{y} \cdot \mathbf{e}_{1}\right)}{\left\|\mathbf{e}_{1}\right\|^{2}}+\cdots+\frac{\left(\mathbf{x} \cdot \mathbf{e}_{n}\right)\left(\mathbf{y} \cdot \mathbf{e}_{n}\right)}{\left\|\mathbf{e}_{n}\right\|^{2}} $$

Short Answer

Expert verified
The dot product is expressed as the sum: \( \sum_{i=1}^n \frac{(\mathbf{x} \cdot \mathbf{e}_i)(\mathbf{y} \cdot \mathbf{e}_i)}{\|\mathbf{e}_i\|^2} \).

Step by step solution

01

Express Vectors in Terms of Basis

Since \( \{\mathbf{e}_{1}, \ldots, \mathbf{e}_{n}\} \) is an orthogonal basis of \( \mathbb{R}^{n} \), we can express \( \mathbf{x} \) and \( \mathbf{y} \) in terms of this basis: \[ \mathbf{x} = a_1\mathbf{e}_1 + a_2\mathbf{e}_2 + \cdots + a_n\mathbf{e}_n \] \[ \mathbf{y} = b_1\mathbf{e}_1 + b_2\mathbf{e}_2 + \cdots + b_n\mathbf{e}_n \] where \( a_i = \frac{\mathbf{x} \cdot \mathbf{e}_i}{\|\mathbf{e}_i\|^2} \) and \( b_i = \frac{\mathbf{y} \cdot \mathbf{e}_i}{\|\mathbf{e}_i\|^2} \).
02

Compute Dot Product Using Expansions

Substitute the expressions for \( \mathbf{x} \) and \( \mathbf{y} \) into the dot product formula. This yields: \[ \mathbf{x} \cdot \mathbf{y} = (a_1\mathbf{e}_1 + a_2\mathbf{e}_2 + \cdots + a_n\mathbf{e}_n) \cdot (b_1\mathbf{e}_1 + b_2\mathbf{e}_2 + \cdots + b_n\mathbf{e}_n) \] By distributivity: \[ \mathbf{x} \cdot \mathbf{y} = \sum_{i=1}^n a_i b_i \mathbf{e}_i \cdot \mathbf{e}_i \]
03

Simplify Using Orthogonality

Due to the orthogonality of the basis, \( \mathbf{e}_i \cdot \mathbf{e}_i = \|\mathbf{e}_i\|^2 \) and \( \mathbf{e}_i \cdot \mathbf{e}_j = 0 \) for \( i eq j \). Thus: \[ \mathbf{x} \cdot \mathbf{y} = \sum_{i=1}^n a_i b_i \|\mathbf{e}_i\|^2 \] Substitute \( a_i = \frac{\mathbf{x} \cdot \mathbf{e}_i}{\|\mathbf{e}_i\|^2} \) and \( b_i = \frac{\mathbf{y} \cdot \mathbf{e}_i}{\|\mathbf{e}_i\|^2} \) back into the sum: \[ \mathbf{x} \cdot \mathbf{y} = \sum_{i=1}^n \frac{(\mathbf{x} \cdot \mathbf{e}_i)(\mathbf{y} \cdot \mathbf{e}_i)}{\|\mathbf{e}_i\|^2} \]
04

Conclusion

The expression for the dot product \( \mathbf{x} \cdot \mathbf{y} \) in terms of the orthogonal basis is verified and matches the given statement of the problem.

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Key Concepts

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

Dot Product
The dot product is a fundamental operation for vectors in mathematics. It helps us to determine the relationship between two vectors. When you compute the dot product of two vectors \(\mathbf{x}\) and \(\mathbf{y}\), you multiply corresponding components and sum their products. This operation is a measure of how much one vector extends in the direction of another. There are key things to remember about the dot product:

  • It results in a scalar value, not a vector.
  • It’s based on the angle between the two vectors. If the angle is zero, the dot product is maximized.
  • If vectors are orthogonal, their dot product is zero.
The dot product is crucial in many areas such as physics, computer graphics, and machine learning, because it provides an efficient way to calculate the angle and projection of vectors.
Orthogonality
Orthogonality is a very important concept in vector spaces. Two vectors are orthogonal if their dot product is zero. In a geometric sense, this means the vectors are perpendicular to each other. This property can simplify many calculations in linear algebra due to its special nature. Here are some points to understand orthogonality better:

  • Orthogonality implies independence; orthogonal vectors do not influence each other.
  • This property is used in constructing orthogonal bases, which simplify computations.
  • Orthogonal vectors have zero projection on each other.
Creating an orthogonal basis from a vector space ensures each vector in the basis is both unique and independent in terms of their directions. It makes working with those vector spaces more manageable and efficient.
Vector Spaces
Vector spaces are basic yet powerful frameworks within which vector geometry and calculus are built. A vector space consists of vectors that can be added together and multiplied by scalars. It’s like a playground for exploring linear relationships, combinations, and transformations.

Some key characteristics of vector spaces include:
  • Vectors can be freely added or scaled within the space.
  • The space includes the zero vector, acting as the identity for addition.
  • There are subspaces, which are subsets of vector spaces satisfying vector space properties.
Understanding vector spaces is crucial because they provide the foundation for many algebraic concepts, like solving linear equations, performing transformations, and more. They offer a way to work systematically with potentially unlimited dimensions, making them invaluable in fields such as physics, computer science, and engineering.

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Most popular questions from this chapter

We often write vectors in \(\mathbb{R}^{n}\) as rows. In each case determine if \(\mathbf{x}\) lies in \(U=\) \(\operatorname{span}\\{\mathbf{y}, \mathbf{z}\\} .\) If \(\mathbf{x}\) is in \(U,\) write it as a linear combination of \(\mathbf{y}\) and \(\mathbf{z}\); if \(\mathbf{x}\) is not in \(U\), show why not. a. \(\mathbf{x}=(2,-1,0,1), \mathbf{y}=(1,0,0,1),\) and \(\mathbf{z}=(0,1,0,1)\) b. \(\mathbf{x}=(1,2,15,11), \mathbf{y}=(2,-1,0,2),\) and \(\quad \mathbf{z}=(1,-1,-3,1)\) c. \(\mathbf{x}=(8,3,-13,20), \mathbf{y}=(2,1,-3,5),\) and \(\mathbf{z}=(-1,0,2,-3)\) d. \(\mathbf{x}=(2,5,8,3), \mathbf{y}=(2,-1,0,5),\) and \(\mathbf{z}=(-1,2,2,-3)\)

If \(A\) is \(m \times n\) of rank \(r,\) show that \(A\) can be factored as \(A=P Q\) where \(P\) is \(m \times r\) with \(r\) independent columns, and \(Q\) is \(r \times n\) with \(r\) independent rows. [Hint: Let \(U A V=\left[\begin{array}{cc}I_{r} & 0 \\ 0 & 0\end{array}\right]\) by Theorem 2.5.3, and write \(U^{-1}=\left[\begin{array}{ll}U_{1} & U_{2} \\ U_{3} & U_{4}\end{array}\right]\) and \(V^{-1}=\left[\begin{array}{ll}V_{1} & V_{2} \\\ V_{3} & V_{4}\end{array}\right]\) in block form, where \(U_{1}\) and \(V_{1}\) are \(\left.r \times r .\right]\)

The yield \(y\) of wheat in bushels per acre appears to be a linear function of the number of days \(x_{1}\) of sunshine, the number of inches \(x_{2}\) of rain, and the number of pounds \(x_{3}\) of fertilizer applied per acre. Find the best fit to the data in the table by an equation of the form \(y=r_{0}+r_{1} x_{1}+r_{2} x_{2}+r_{3} x_{3} .\) [Hint: If a calculator for inverting \(A^{T} A\) is not available, the inverse is given in the answer.] $$ \begin{array}{|c|c|c|c|} \hline y & x_{1} & x_{2} & x_{3} \\ \hline 28 & 50 & 18 & 10 \\ 30 & 40 & 20 & 16 \\ 21 & 35 & 14 & 10 \\ 23 & 40 & 12 & 12 \\ 23 & 30 & 16 & 14 \\ \hline \end{array} $$

Show that the only matrix similar to a scalar matrix \(A=r I, r\) in \(\mathbb{R},\) is \(A\) itself.

In each case either show that the statement is true or give an example showing that it is false. a. Every independent set in \(\mathbb{R}^{n}\) is orthogonal. b. If \(\\{\mathbf{x}, \mathbf{y}\\}\) is an orthogonal set in \(\mathbb{R}^{n},\) then \(\\{\mathbf{x}, \mathbf{x}+\mathbf{y}\\}\) is also orthogonal. c. If \(\\{\mathbf{x}, \mathbf{y}\\}\) and \(\\{\mathbf{z}, \mathbf{w}\\}\) are both orthogonal in \(\mathbb{R}^{n}\), then \(\\{\mathbf{x}, \mathbf{y}, \mathbf{z}, \mathbf{w}\\}\) is also orthogonal. d. If \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}\right\\}\) and \(\left\\{\mathbf{y}_{1}, \mathbf{y}_{2}, \mathbf{y}_{3}\right\\}\) are both or- thogonal and \(\mathbf{x}_{i} \cdot \mathbf{y}_{j}=0\) for all \(i\) and \(j,\) then \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{y}_{1}, \mathbf{y}_{2}, \mathbf{y}_{3}\right\\}\) is orthogonal. e. If \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{n}\right\\}\) is orthogonal in \(\mathbb{R}^{n}\), then \(\mathbb{R}^{n}=\operatorname{span}\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{n}\right\\}\) f. If \(\mathbf{x} \neq \mathbf{0}\) in \(\mathbb{R}^{n},\) then \(\\{\mathbf{x}\\}\) is an orthogonal set.

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