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In Exercises \(3-6,\) verify that \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}\right\\}\) is an orthogonal set, and then find the orthogonal projection of \(\mathbf{y}\) onto \(\operatorname{Span}\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}\right\\}\) $$ \mathbf{y}=\left[\begin{array}{r}{-1} \\ {4} \\ {3}\end{array}\right], \mathbf{u}_{1}=\left[\begin{array}{l}{1} \\ {1} \\ {0}\end{array}\right], \mathbf{u}_{2}=\left[\begin{array}{r}{-1} \\ {1} \\ {0}\end{array}\right] $$

Short Answer

Expert verified
The orthogonal projection of \( \mathbf{y} \) onto \( \text{Span}(\{ \mathbf{u}_1, \mathbf{u}_2 \}) \) is \( \begin{bmatrix} -1 \\ 4 \\ 0 \end{bmatrix} \).

Step by step solution

01

Check Orthogonality

To check if the set \( \{ \mathbf{u}_1, \mathbf{u}_2 \} \) is orthogonal, we calculate the dot product \( \mathbf{u}_1 \cdot \mathbf{u}_2 \). For \( \mathbf{u}_1 = \begin{bmatrix} 1 \ 1 \ 0 \end{bmatrix} \) and \( \mathbf{u}_2 = \begin{bmatrix} -1 \ 1 \ 0 \end{bmatrix} \), the dot product is \( 1 \times (-1) + 1 \times 1 + 0 \times 0 = -1 + 1 = 0 \). Since the dot product is zero, \( \{ \mathbf{u}_1, \mathbf{u}_2 \} \) is an orthogonal set.
02

Find Projection onto u1

The projection of \( \mathbf{y} \) onto \( \mathbf{u}_1 \) is given by \( \text{proj}_{\mathbf{u}_1}(\mathbf{y}) = \frac{\mathbf{y} \cdot \mathbf{u}_1}{\mathbf{u}_1 \cdot \mathbf{u}_1} \mathbf{u}_1 \). Calculate \( \mathbf{y} \cdot \mathbf{u}_1 = (-1) \cdot 1 + 4 \cdot 1 + 3 \cdot 0 = 3 \). Then compute \( \mathbf{u}_1 \cdot \mathbf{u}_1 = 1^2 + 1^2 + 0^2 = 2 \). Now, \( \text{proj}_{\mathbf{u}_1}(\mathbf{y}) = \frac{3}{2} \begin{bmatrix} 1 \ 1 \ 0 \end{bmatrix} = \begin{bmatrix} \frac{3}{2} \ \frac{3}{2} \ 0 \end{bmatrix} \).
03

Find Projection onto u2

The projection of \( \mathbf{y} \) onto \( \mathbf{u}_2 \) is \( \text{proj}_{\mathbf{u}_2}(\mathbf{y}) = \frac{\mathbf{y} \cdot \mathbf{u}_2}{\mathbf{u}_2 \cdot \mathbf{u}_2} \mathbf{u}_2 \). Calculate \( \mathbf{y} \cdot \mathbf{u}_2 = (-1) \times (-1) + 4 \times 1 + 3 \times 0 = 5 \). Then \( \mathbf{u}_2 \cdot \mathbf{u}_2 = (-1)^2 + 1^2 + 0^2 = 2 \). Thus, \( \text{proj}_{\mathbf{u}_2}(\mathbf{y}) = \frac{5}{2} \begin{bmatrix} -1 \ 1 \ 0 \end{bmatrix} = \begin{bmatrix} -\frac{5}{2} \ \frac{5}{2} \ 0 \end{bmatrix} \).
04

Calculate Total Projection

The total projection of \( \mathbf{y} \) onto the span of \( \{ \mathbf{u}_1, \mathbf{u}_2 \} \) is the sum of the individual projections. So, \( \text{proj}_{\text{Span}(\mathbf{u}_1, \mathbf{u}_2)}(\mathbf{y}) = \text{proj}_{\mathbf{u}_1}(\mathbf{y}) + \text{proj}_{\mathbf{u}_2}(\mathbf{y}) = \begin{bmatrix} \frac{3}{2} \ \frac{3}{2} \ 0 \end{bmatrix} + \begin{bmatrix} -\frac{5}{2} \ \frac{5}{2} \ 0 \end{bmatrix} = \begin{bmatrix} -1 \ 4 \ 0 \end{bmatrix} \).

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Key Concepts

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

Orthogonal Projection
When dealing with vectors, the concept of orthogonal projection is crucial for understanding how a vector can be represented within a subspace. The orthogonal projection of a vector \(\mathbf{y}\) onto another vector \(\mathbf{u}\) is essentially the component of \(\mathbf{y}\) that points in the direction of \(\mathbf{u}\). This is especially useful in finding how much of \(\mathbf{y}\) can be described within the span of \(\mathbf{u}\) or a set of basis vectors.
To compute the orthogonal projection of \(\mathbf{y}\) onto a vector \(\mathbf{u}\), the formula is given by:
  • \( \text{proj}_{\mathbf{u}}(\mathbf{y}) = \frac{\mathbf{y} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u} \)
This formula works because it scales the unit vector in the direction of \(\mathbf{u}\) by the amount of \(\mathbf{y}\) that is "projected" onto \(\mathbf{u}\). The result is a vector that lies directly along \(\mathbf{u}\).
An important aspect of orthogonal projection in multi-dimensional space, particularly with orthogonal sets, is ensuring that none of the original vector's full magnitude is lost in the process of breaking it down into components within the span of the vectors involved.
Dot Product
The dot product is a fundamental operation in vector algebra that helps determine the relationship between two vectors. It is a scalar value that gives insight into the angle between vectors and their respective magnitudes.
Calculated as:
  • \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 + \ldots \)

If the dot product of two vectors is zero, this signifies that the vectors are orthogonal, meaning they are at right angles to each other in the vector space.
For our example, checking the dot product of \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \) gives zero, confirming they are orthogonal. This orthogonality plays a key role in simplifying calculations for projections and for verifying the integrity of vector sets in higher-dimensional analysis. When working with an orthogonal set, each vector operates independently, minimizing complications when translating or synthesizing vector data.
Span of Vectors
The span of a set of vectors refers to all possible vectors that can be constructed as linear combinations of the set. This is an essential concept in understanding vector spaces because it defines the space created by the combination of multiple vectors.
In mathematical terms, given vectors \(\mathbf{u}_1\) and \(\mathbf{u}_2\), the span is defined as:
  • \( \text{Span}\{\mathbf{u}_1, \mathbf{u}_2\} = \{c_1 \mathbf{u}_1 + c_2 \mathbf{u}_2 \mid c_1, c_2 \in \mathbb{R}\} \)

This means any vector in the space described by \(\mathbf{u}_1\) and \(\mathbf{u}_2\) can be expressed as a weighted sum of these vectors, where \(c_1\) and \(c_2\) are real numbers. In our exercise case, the span of \(\mathbf{u}_1\) and \(\mathbf{u}_2\) encapsulates a plane in the three-dimensional space.
Calculating the projection of a vector like \(\mathbf{y}\) onto this span thereby gives a clear representation of how much of \(\mathbf{y}\) can lie within this plane, effectively "projecting" it into the defined space. This is fundamental when solving systems of equations or optimizing functions, as it allows you to simplify complex three-dimensional problems into two dimensions.

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Most popular questions from this chapter

A simple curve that often makes a good model for the variable costs of a company, as a function of the sales level \(x\), has the form \(y=\beta_{1} x+\beta_{2} x^{2}+\beta_{3} x^{3} .\) There is no constant term because fixed costs are not included. a. Give the design matrix and the parameter vector for the linear model that leads to a least-squares fit of the equation above, with data \(\left(x_{1}, y_{1}\right), \ldots,\left(x_{n}, y_{n}\right)\) b. [M] Find the least-squares curve of the form above to fit the data \((4,1.58),(6,2.08),(8,2.5),(10,2.8),(12,3.1)\) \((14,3.4),(16,3.8),\) and \((18,4.32),\) with values in thou- sands. If possible, produce a graph that shows the data points and the graph of the cubic approximation.

In Exercises 17 and \(18,\) all vectors and subspaces are in \(\mathbb{R}^{n} .\) Mark each statement True or False. Justify each answer. a. If \(W=\operatorname{Span}\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}\right\\}\) with \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}\right\\}\) linearly independent, and if \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\\}\) is an orthogonal set in \(W,\) then \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\\}\) is a basis for \(W\) b. If \(\mathbf{x}\) is not in a subspace \(W,\) then \(\mathbf{x}-\operatorname{proj}_{W} \mathbf{x}\) is not zero. c. In a QR factorization, say \(A=Q R\) (when \(A\) has lincarly independent columns), the columns of \(Q\) form an orthonormal basis for the column space of \(A .\)

Let \(W=\operatorname{Span}\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\} .\) Show that if \(\mathbf{x}\) is orthogonal to each \(\mathbf{v}_{j},\) for \(1 \leq j \leq p,\) then \(\mathbf{x}\) is orthogonal to every vector in \(W .\)

In Exercises 11 and \(12,\) find the closest point to \(y\) in the subspace \(W\) spanned by \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2} .\) $$ \mathbf{y}=\left[\begin{array}{l}{3} \\ {1} \\ {5} \\\ {1}\end{array}\right], \mathbf{v}_{1}=\left[\begin{array}{r}{3} \\ {1} \\\ {-1} \\ {1}\end{array}\right], \mathbf{v}_{2}=\left[\begin{array}{r}{1} \\\ {-1} \\ {1} \\ {-1}\end{array}\right] $$

In Exercises \(7-10,\) let \(W\) be the subspace spanned by the \(\mathbf{u}^{\prime}\) 's, and write \(\mathbf{y}\) as the sum of a vector in \(W\) and a vector orthogonal to \(W\) $$ \mathbf{y}=\left[\begin{array}{l}{3} \\ {4} \\ {5} \\\ {6}\end{array}\right], \mathbf{u}_{1}=\left[\begin{array}{r}{1} \\ {1} \\\ {0} \\ {-1}\end{array}\right], \mathbf{u}_{2}=\left[\begin{array}{l}{1} \\\ {0} \\ {1} \\ {1}\end{array}\right], \mathbf{u}_{3}=\left[\begin{array}{r}{0} \\ {-1} \\ {1} \\\ {-1}\end{array}\right] $$

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