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Let \(\mathbf{x}=(1,1,1,1)^{T}\) and \(\mathbf{y}=(8,2,2,0)^{T}\) (a) Determine the angle \(\theta\) between \(\mathbf{x}\) and \(\mathbf{y}\) (b) Find the vector projection \(\mathbf{p}\) of \(\mathbf{x}\) onto \(\mathbf{y}\) (c) Verify that \(\mathbf{x}-\mathbf{p}\) is orthogonal to \(\mathbf{p}\) (d) Compute \(\|\mathbf{x}-\mathbf{p}\|_{2},\|\mathbf{p}\|_{2},\|\mathbf{x}\|_{2}\) and verify that the Pythagorean law is satisfied.

Short Answer

Expert verified
The angle between vectors \(\mathbf{x}\) and \(\mathbf{y}\) is \(\theta = \arccos\left(\frac{12}{2\sqrt{72}}\right)\). The vector projection of \(\mathbf{x}\) onto \(\mathbf{y}\) is \(\mathbf{p} = \left(\frac{4}{3}, \frac{1}{3}, \frac{1}{3}, 0\right)\). It is verified that \(\mathbf{x}-\mathbf{p}\) is orthogonal to \(\mathbf{p}\). The norms computed are \(\|\mathbf{x} - \mathbf{p}\|_{2} = \sqrt{\frac{20}{9}}\), \(\|\mathbf{p}\|_{2} = \sqrt{\frac{16}{9}}\), and \(\|\mathbf{x}\|_{2} = 2\). The Pythagorean law is satisfied as \((\|\mathbf{x} - \mathbf{p}\|_{2})^{2} + (\|\mathbf{p}\|_{2})^{2} = (\|\mathbf{x}\|_{2})^{2}\).

Step by step solution

01

(a) Finding the angle between \(\mathbf{x}\) and \(\mathbf{y}\)

First, calculate the dot product of \(\mathbf{x}\) and \(\mathbf{y}\): \(\mathbf{x} \cdot \mathbf{y} = (1)(8) + (1)(2) + (1)(2) + (1)(0) = 12\) Next, find the magnitudes of \(\mathbf{x}\) and \(\mathbf{y}\): \(\|\mathbf{x}\| = \sqrt{(1)^{2} + (1)^{2} + (1)^{2} + (1)^{2}} = \sqrt{4} = 2\) \(\|\mathbf{y}\| = \sqrt{(8)^{2} + (2)^{2} + (2)^{2} + (0)^{2}} = \sqrt{72}\) Now, we can find the angle \(\theta\) using the dot product formula: \(\cos(\theta) = \frac{\mathbf{x} \cdot \mathbf{y}}{\|\mathbf{x}\|\|\mathbf{y}\|} = \frac{12}{2 \sqrt{72}}\) \(\theta= \arccos\left(\frac{12}{2\sqrt{72}}\right)\)
02

(b) Finding the vector projection of \(\mathbf{x}\) onto \(\mathbf{y}\)

To find the vector projection, we first need to calculate the scalar projection: \(\frac{\mathbf{x} \cdot \mathbf{y}}{\|\mathbf{y}\|^{2}}=\frac{12}{72}=\frac{1}{6}\) Now, we can find the vector projection \(\mathbf{p}\): \(\mathbf{p} = \frac{\mathbf{x} \cdot \mathbf{y}}{\|\mathbf{y}\|^{2}} \mathbf{y} = \frac{1}{6} (8, 2, 2, 0) = \left(\frac{4}{3}, \frac{1}{3}, \frac{1}{3}, 0\right)\)
03

(c) Verifying that \(\mathbf{x} - \mathbf{p}\) is orthogonal to \(\mathbf{p}\)

To prove that \(\mathbf{x} - \mathbf{p}\) is orthogonal to \(\mathbf{p}\), their dot product needs to be zero: \((\mathbf{x} - \mathbf{p}) \cdot \mathbf{p} = \left(1 - \frac{4}{3}, 1 - \frac{1}{3}, 1 - \frac{1}{3}, 1\right) \cdot \left(\frac{4}{3}, \frac{1}{3}, \frac{1}{3}, 0\right) = 0\) This proves that \(\mathbf{x} - \mathbf{p}\) is orthogonal to \(\mathbf{p}\).
04

(d) Verifying the Pythagorean law

First, compute the required norms: \(\|\mathbf{x} - \mathbf{p}\|_{2} = \left\|\left(\frac{-1}{3}, \frac{2}{3}, \frac{2}{3}, 1\right)\right\|_{2} = \sqrt{\left(\frac{-1}{3}\right)^{2} + \left(\frac{2}{3}\right)^{2} + \left(\frac{2}{3}\right)^{2} + 1^{2}} = \sqrt{\frac{20}{9}}\) \(\|\mathbf{p}\|_{2} =\left(\frac{4}{3}, \frac{1}{3}, \frac{1}{3}, 0\right)= \sqrt{\left(\frac{4}{3}\right)^{2} + \left(\frac{1}{3}\right)^{2} + \left(\frac{1}{3}\right)^{2}+0^{2}}= \sqrt{\frac{16}{9}}\) \(\|\mathbf{x}\|_{2} = \| (1, 1, 1, 1) \|_{2} = \sqrt{1^{2} + 1^{2} + 1^{2} + 1^{2}}= 2\) Now we verify if the Pythagorean law is satisfied: \((\|\mathbf{x} - \mathbf{p}\|_{2})^{2} + (\|\mathbf{p}\|_{2})^{2} = \frac{20}{9} + \frac{16}{9} = \frac{36}{9} = 4\) \((\|\mathbf{x}\|_{2})^{2} = 2^{2} = 4\) This proves that the Pythagorean law holds: \((\|\mathbf{x} - \mathbf{p}\|_{2})^{2} + (\|\mathbf{p}\|_{2})^{2} = (\|\mathbf{x}\|_{2})^{2}\).

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Key Concepts

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

Angle Between Vectors
In vector calculus, the angle between two vectors can be found using the dot product formula. Given vectors \(\mathbf{x}\) and \(\mathbf{y}\), if we know their dot product and respective magnitudes, the angle \(\theta\) between them can be computed.
The cosine of the angle is given by:
  • \(\cos(\theta) = \frac{\mathbf{x} \cdot \mathbf{y}}{\|\mathbf{x}\| \|\mathbf{y}\|}\)
For instance, in the exercise, we've used this to compute \(\cos(\theta) = \frac{12}{2\sqrt{72}}\). This formula derives from relating the dot product to the cosine of the angle between the vectors.
Understanding this relationship can help determine how similar directions of the two vectors are. If the cosine is 1, they point in the same direction; if it is -1, they are opposite; and if it is 0, they are orthogonal.
Orthogonal Vectors
Orthogonal vectors are vectors that are perpendicular to each other. Mathematically, two vectors \(\mathbf{a}\) and \(\mathbf{b}\) are orthogonal if their dot product is zero, i.e., \(\mathbf{a} \cdot \mathbf{b} = 0\).
This concept plays a crucial role in determining if a certain transformation or operation on vectors maintains their orthogonality.
In the exercise, we verify this condition by computing the dot product of \(\mathbf{x} - \mathbf{p}\) with \(\mathbf{p}\). It results in zero, confirming that these vectors are orthogonal.
This verification involves simple calculations, but it has important implications, particularly in projections and vector spaces. Demonstrating orthogonality ensures that the vector \(\mathbf{x} - \mathbf{p}\) lies entirely within the plane perpendicular to \(\mathbf{p}\), which is crucial when managing decompositions in vector spaces.
Pythagorean Theorem in Vector Spaces
The Pythagorean theorem in the context of vector spaces extends its classic form to involve vector magnitudes and orthogonality. If \(\mathbf{u}\) and \(\mathbf{v}\) are orthogonal vectors, the theorem states:
  • \(\|\mathbf{u} + \mathbf{v}\|^2 = \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2\)
Essentially, this theorem in vector spaces is used to affirm the equality of the sums of squares of orthogonal components to the square of the resultant vector.
In the exercise, after confirming \(\mathbf{x} - \mathbf{p}\) is orthogonal to \(\mathbf{p}\), we use this principle to verify the Pythagorean identity. By checking if the equation \((\|\mathbf{x} - \mathbf{p}\|_{2})^{2} + (\|\mathbf{p}\|_{2})^{2} = (\|\mathbf{x}\|_{2})^{2}\) is satisfied, it adds integrity to our vector decomposition and projections.
Hence, confirming adherence to the theorem is essential in tasks that require breaking down vectors into orthogonal components, ensuring the calculations are consistent across the vector space.
Dot Product of Vectors
The dot product (or scalar product) of two vectors is a central operation in vector algebra. It provides valuable information like the angle between vectors and can determine their orthogonality.
Algebraically, the dot product of vectors \(\mathbf{a} = (a_1, a_2, ..., a_n)\) and \(\mathbf{b} = (b_1, b_2, ..., b_n)\) is calculated as:
  • \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + ... + a_nb_n\)
This pseudoscientific formula reveals combinations of numbers to show how similar two vectors are on an axis. It leads directly to calculating projections and angles between vectors.
In the exercise, the dot product of \(\mathbf{x}\) and \(\mathbf{y}\) is 12, obtained from summing the products of corresponding components, signifying the overlap and orientations relative to each other.
Understanding and using the dot product allows students to solve problems related to projections, forces, and finding unknown angles, making it a versatile tool in various fields of science and engineering.

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Most popular questions from this chapter

Let \(A\) be an \(m \times 2\) matrix. Show that if both the classical Gram- Schmidt process and the modified Gram-Schmidt process are applied to the column vectors of \(A,\) then both algorithms will produce the exact same QR factorization, even when the computations are carried out in finite-precision arithmetic (i.e., show that both algorithms will perform the exact same arithmetic computations).

Let \(S\) be a subspace of an inner product space \(V\) Let \(\left\\{\mathbf{x}_{1}, \ldots, \mathbf{x}_{n}\right\\}\) be an orthogonal basis for \(S\) and let \(\mathbf{x} \in V .\) Show that the best least squares approximation to \(\mathbf{x}\) by elements of \(S\) is given by $$\mathbf{p}=\sum_{i=1}^{n} \frac{\left\langle\mathbf{x}, \mathbf{x}_{i}\right\rangle}{\left\langle\mathbf{x}_{i}, \mathbf{x}_{i}\right\rangle} \mathbf{x}_{i}$$

If \(Q\) is an \(n \times n\) orthogonal matrix and \(\mathbf{x}\) and y are nonzero vectors in \(\mathbb{R}^{n},\) then how does the angle between \(Q \mathbf{x}\) and \(Q \mathbf{y}\) compare with the angle between \(\mathbf{x}\) and \(\mathbf{y} ?\) Prove your answer.

Consider the vector space \(\mathbb{R}^{n}\) with inner product \(\langle\mathbf{x}, \mathbf{y}\rangle=\mathbf{x}^{T} \mathbf{y} .\) Show that for any \(n \times n\) matrix \(A\) (a) \(\langle A \mathbf{x}, \mathbf{y}\rangle=\left\langle\mathbf{x}, A^{T} \mathbf{y}\right\rangle\) (b) \(\left\langle A^{T} A \mathbf{x}, \mathbf{x}\right\rangle=\|A \mathbf{x}\|^{2}\)

Let \(A\) be an \(m \times 3\) matrix. Let \(Q R\) be the QR factorization obtained when the classical Gram-Schmidt process is applied to the column vectors of \(A,\) and let \(\tilde{Q} \tilde{R}\) be the factorization obtained when the modified Gram-Schmidt process is used. Show that if all computations were carried out using exact arithmetic then we would have $$\tilde{Q}=Q \quad \text { and } \quad \tilde{R}=R$$ and show that when the computations are done in finite-precision arithmetic, \(\tilde{r}_{23}\) will not necessarily be equal to \(r_{23}\) and consequently \(\tilde{r}_{33}\) and \(\tilde{\mathbf{q}}_{3}\) will not necessarily be the same as \(r_{33}\) and \(\mathbf{q}_{3}\)

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