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Chapter 7: Problem 2

In Exercises 1 and \(2,\) let \(\mathbf{u}=\left[\begin{array}{r}2 \\\ -1\end{array}\right]\) and \(\mathbf{v}=\left[\begin{array}{l}3 \\\ 4\end{array}\right] .\) Compute \((a)\langle\mathbf{u}, \mathbf{v}\rangle\) \((b)\|\mathbf{u}\|\) \((c) \mathrm{d}(\mathbf{u}, \mathbf{v})\) \(\langle\mathbf{u}, \mathbf{v}\rangle\) is the inner product of Example 7.3 with \(A=\) \(\left[\begin{array}{rr}4 & -2 \\ -2 & 7\end{array}\right]\)

Short Answer

Expert verified
(a) \(-14\); (b) \(\sqrt{31}\); (c) \(\sqrt{159}\).

Step by step solution

01

Compute the Inner Product

First, compute the inner product \( \langle \mathbf{u}, \mathbf{v} \rangle \) using matrix \( A \). The inner product is given by \( \langle \mathbf{u}, \mathbf{v} \rangle = \mathbf{u}^T A \mathbf{v} \).\\\( \mathbf{u} = \begin{bmatrix} 2 \ -1 \end{bmatrix}, \quad \mathbf{v} = \begin{bmatrix} 3 \ 4 \end{bmatrix}, \quad A = \begin{bmatrix} 4 & -2 \ -2 & 7 \end{bmatrix} \)\\Calculate \( A \mathbf{v} \):\\[ A \mathbf{v} = \begin{bmatrix} 4 & -2 \ -2 & 7 \end{bmatrix} \begin{bmatrix} 3 \ 4 \end{bmatrix} = \begin{bmatrix} 4(3) - 2(4) \ -2(3) + 7(4) \end{bmatrix} = \begin{bmatrix} 12 - 8 \ -6 + 28 \end{bmatrix} = \begin{bmatrix} 4 \ 22 \end{bmatrix} \] \\Then, calculate \( \mathbf{u}^T \cdot \begin{bmatrix} 4 \ 22 \end{bmatrix} \):\\[ \langle \mathbf{u}, \mathbf{v} \rangle = \begin{bmatrix} 2 & -1 \end{bmatrix} \begin{bmatrix} 4 \ 22 \end{bmatrix} = 2(4) + (-1)(22) = 8 - 22 = -14 \]
02

Compute the Norm of \( \mathbf{u} \)

The norm \( \| \mathbf{u} \| \) is given by \( \| \mathbf{u} \| = \sqrt{\mathbf{u}^T A \mathbf{u}} \).\\Compute \( A \mathbf{u} \):\\[ A \mathbf{u} = \begin{bmatrix} 4 & -2 \ -2 & 7 \end{bmatrix} \begin{bmatrix} 2 \ -1 \end{bmatrix} = \begin{bmatrix} 4(2) - 2(-1) \ -2(2) + 7(-1) \end{bmatrix} = \begin{bmatrix} 8 + 2 \ -4 - 7 \end{bmatrix} = \begin{bmatrix} 10 \ -11 \end{bmatrix} \] \\Then, calculate \( \mathbf{u}^T \cdot \begin{bmatrix} 10 \ -11 \end{bmatrix} \):\\[ \| \mathbf{u} \| = \sqrt{\begin{bmatrix} 2 & -1 \end{bmatrix} \begin{bmatrix} 10 \ -11 \end{bmatrix}} = \sqrt{2(10) + (-1)(-11)} = \sqrt{20 + 11} = \sqrt{31} \] \approx 5.57
03

Compute the Distance \( d(\mathbf{u}, \mathbf{v}) \)

The distance between vectors is determined by \( d(\mathbf{u}, \mathbf{v}) = \sqrt{(\mathbf{u} - \mathbf{v})^T A (\mathbf{u} - \mathbf{v})} \).\\First, find \( \mathbf{u} - \mathbf{v} \):\\[ \mathbf{u} - \mathbf{v} = \begin{bmatrix} 2 \ -1 \end{bmatrix} - \begin{bmatrix} 3 \ 4 \end{bmatrix} = \begin{bmatrix} 2 - 3 \ -1 - 4 \end{bmatrix} = \begin{bmatrix} -1 \ -5 \end{bmatrix} \] \\Next, calculate \( A (\mathbf{u} - \mathbf{v}) \):\\[ A \begin{bmatrix} -1 \ -5 \end{bmatrix} = \begin{bmatrix} 4 & -2 \ -2 & 7 \end{bmatrix} \begin{bmatrix} -1 \ -5 \end{bmatrix} = \begin{bmatrix} 4(-1) - 2(-5) \ -2(-1) + 7(-5) \end{bmatrix} = \begin{bmatrix} -4 + 10 \ 2 - 35 \end{bmatrix} = \begin{bmatrix} 6 \ -33 \end{bmatrix} \] \\Finally, calculate \( (\mathbf{u} - \mathbf{v})^T \cdot \begin{bmatrix} 6 \ -33 \end{bmatrix} \):\\[ d(\mathbf{u}, \mathbf{v}) = \sqrt{\begin{bmatrix} -1 & -5 \end{bmatrix} \begin{bmatrix} 6 \ -33 \end{bmatrix}} = \sqrt{(-1)(6) + (-5)(-33)} = \sqrt{-6 + 165} = \sqrt{159} \] \approx 12.61

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Key Concepts

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

Inner Product
The inner product is a fundamental concept in vector spaces. It allows us to compute an angle-like relationship between two vectors. For vectors \(\mathbf{u}\) and \(\mathbf{v}\), the inner product \(\langle \mathbf{u}, \mathbf{v} \rangle\) is often calculated with the formula \(\mathbf{u}^T A \mathbf{v}\), where \( A \) is a symmetric matrix representing the mode of measurement in the given space.

  • It combines both vectors and the matrix to obtain a scalar.
  • The inner product can show orthogonality: if \(\langle \mathbf{u}, \mathbf{v} \rangle = 0\), the vectors are orthogonal.
  • It assesses similarity. A high value indicates higher similarity.
  • Applications extend to areas like physics and geometry for computing projections.

In our example, computing the inner product involved taking the transpose of \(\mathbf{u}\), multiplying with matrix \( A \), then with vector \(\mathbf{v}\). The result, \(-14\), suggests a certain angle-related dissociation between them in the vector space modeled by \( A \). Understanding the inner product helps analyze vector alignment and engagement within a given context.
Norm of a Vector
The norm of a vector is essentially its length or magnitude in a given space. It's a measure that shows how long the vector is, often represented as \(\|\mathbf{u}\|\). To find a vector's norm, you square its elements, sum them, and then take the square root of that sum. Specifically, in our exercise, the norm considers the matrix \( A \) to adjust the space representation.

  • Norm can be visualized as distance from the origin in a vector space.
  • It's always non-negative and provides a way to scale vectors uniformly.
  • The Euclidean norm \(\sqrt{\mathbf{u}^T A \mathbf{u}}\) was used here due to matrix presence.
  • It assists in standardizing data, crucial for many algorithmic applications.

In the example given, through a sequence of matrix multiplications, \(\mathbf{u}^T\) was post-multiplied by \( A \mathbf{u}\) and rooted, resulting in \(\sqrt{31} \approx 5.57\). This norm gives us indispensable insights into vector magnitudes, which aids in various calculations including those involving projections and normalizations.
Distance Between Vectors
The distance between vectors provides a quantitative measure of how far apart two vectors are in a vector space. Calculating this distance incorporates the concept of the norm and adjusts for any vector offsets using matrix influence if specified. Expressed with \( \mathrm{d}(\mathbf{u}, \mathbf{v}) = \sqrt{(\mathbf{u} - \mathbf{v})^T A (\mathbf{u} - \mathbf{v})} \), it introduces nuances about their position relative to each other.

  • This metric serves as a fundamental building block in fields like machine learning, where it infers similarity between data points.
  • Useful in clustering and classification problems as it helps to segregate or merge entities.
  • Acts in linear spaces adjusted by matrix \( A \) to scale or skew based on requirements.

In this problem, vector subtraction preceded matrix transformation, leading to a norm operation reflecting integrated distance calculation, \(\sqrt{159} \approx 12.61\). The result illustrates an abstract distance photo of \(\mathbf{u}\) from \(\mathbf{v}\), providing foundational knowledge integral to spatial analysis and engineered approximations.

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Most popular questions from this chapter

Consider the linear system \(A \mathbf{x}=\mathbf{b},\) where \(A\) is invertible. Suppose an error \(\Delta\) b changes \(b\) to \(b^{\prime}=b+\Delta b\). Let \(x^{\prime}\) be the solution to the new system; that is, \(A \mathbf{x}^{\prime}=\mathbf{b}^{\prime} \cdot\) Let \(\mathbf{x}^{\prime}=\mathbf{x}+\Delta \mathbf{x}\) so that \(\Delta \mathbf{x}\) represents the resulting error in the solution of the system. Show that $$\frac{\|\Delta \mathbf{x}\|}{\|\mathbf{x}\|} \leq \operatorname{cond}(A) \frac{\|\Delta \mathbf{b}\|}{\|\mathbf{b}\|}$$ for any compatible matrix norm.

Find the least squares approximating line for the given points and compute the corresponding least squares error. $$(1,1),(2,3),(3,4),(4,5),(5,7)$$

Find the singular values of the given matrix. $$A=\left[\begin{array}{rrr}1 & 0 & 1 \\\0 & -3 & 0 \\\1 & 0 & 1\end{array}\right]$$

Prove that if \(A\) is a positive definite matrix with \(\mathrm{SVD}\) \(A=U \Sigma V^{T},\) then \(U=V\).

Find the standard matrix of the orthogonal projection onto the subspace \(W\). Then use this matrix to find the orthogonal projection of v onto \(W\). $$W=\operatorname{span}\left(\left[\begin{array}{r} 1 \\ -2 \\ 1 \end{array}\right],\left[\begin{array}{r} 1 \\ 0 \\ -1 \end{array}\right]\right), \mathbf{v}=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]$$
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