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Chapter 6: Problem 5

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 \(\operatorname{Span}\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}\right\\}\) $$ \mathbf{y}=\left[\begin{array}{r}{-1} \\ {2} \\ {6}\end{array}\right], \mathbf{u}_{1}=\left[\begin{array}{r}{3} \\ {-1} \\ {2}\end{array}\right], \mathbf{u}_{2}=\left[\begin{array}{r}{1} \\ {-1} \\ {-2}\end{array}\right] $$

Short Answer

Expert verified
The orthogonal projection of \(\mathbf{y}\) is \([-1, 2, 6]\).

Step by step solution

01

Verify Orthogonality

To confirm that \(\{ \mathbf{u}_1, \mathbf{u}_2 \}\) is an orthogonal set, compute the dot product \(\mathbf{u}_1 \cdot \mathbf{u}_2\).Calculate: \( \mathbf{u}_1 \cdot \mathbf{u}_2 = [3, -1, 2] \cdot [1, -1, -2] = 3\times 1 + (-1)\times (-1) + 2\times (-2) \).Simplifying this gives: \(3 + 1 - 4 = 0\).Since the dot product is zero, \(\{ \mathbf{u}_1, \mathbf{u}_2 \}\) is an orthogonal set.
02

Compute Projection onto Basis Vector \(\mathbf{u}_1\)

To find the projection of \(\mathbf{y}\) onto \(\mathbf{u}_1\), use the formula:\[ \text{proj}_{\mathbf{u}_1}(\mathbf{y}) = \frac{\mathbf{y} \cdot \mathbf{u}_1}{\mathbf{u}_1 \cdot \mathbf{u}_1} \mathbf{u}_1 \]First, calculate \(\mathbf{y} \cdot \mathbf{u}_1\):\( [-1, 2, 6] \cdot [3, -1, 2] = -3 - 2 + 12 = 7 \).Next, calculate \(\mathbf{u}_1 \cdot \mathbf{u}_1\):\( [3, -1, 2] \cdot [3, -1, 2] = 9 + 1 + 4 = 14\).Then, find \( \text{proj}_{\mathbf{u}_1}(\mathbf{y}) = \frac{7}{14} \mathbf{u}_1 = \frac{1}{2} [3, -1, 2] = \left[\frac{3}{2}, -\frac{1}{2}, 1\right] \).
03

Compute Projection onto Basis Vector \(\mathbf{u}_2\)

Now find the projection of \(\mathbf{y}\) onto \(\mathbf{u}_2\) using the formula:\[ \text{proj}_{\mathbf{u}_2}(\mathbf{y}) = \frac{\mathbf{y} \cdot \mathbf{u}_2}{\mathbf{u}_2 \cdot \mathbf{u}_2} \mathbf{u}_2 \]Calculate \(\mathbf{y} \cdot \mathbf{u}_2\):\( [-1, 2, 6] \cdot [1, -1, -2] = -1 - 2 - 12 = -15 \).Next, calculate \(\mathbf{u}_2 \cdot \mathbf{u}_2\):\( [1, -1, -2] \cdot [1, -1, -2] = 1 + 1 + 4 = 6\).Then, find \( \text{proj}_{\mathbf{u}_2}(\mathbf{y}) = \frac{-15}{6} \mathbf{u}_2 = -\frac{5}{2} [1, -1, -2] = \left[-\frac{5}{2}, \frac{5}{2}, 5\right] \).
04

Find Total Orthogonal Projection

Sum the projections from Steps 2 and 3 to find the full orthogonal projection of \(\mathbf{y}\) onto \(\operatorname{Span}\{\mathbf{u}_1, \mathbf{u}_2\}\):\[ \text{proj}_{\operatorname{Span}\{\mathbf{u}_1, \mathbf{u}_2\}}(\mathbf{y}) = \text{proj}_{\mathbf{u}_1}(\mathbf{y}) + \text{proj}_{\mathbf{u}_2}(\mathbf{y}) \]Substitute the values:\[ \left[\frac{3}{2}, -\frac{1}{2}, 1\right] + \left[-\frac{5}{2}, \frac{5}{2}, 5\right] = \left[-1, 2, 6\right] \]Thus, the orthogonal projection of \(\mathbf{y}\) is \([-1, 2, 6] \).

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Key Concepts

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

Orthogonal Sets
An orthogonal set of vectors in linear algebra is one where each pair of different vectors is perpendicular to each other. This means when you calculate the dot product between any two vectors in the set, the result is zero. Looking at the vectors \(\{\mathbf{u}_1, \mathbf{u}_2\}\), to check if they form an orthogonal set, we performed the dot product calculation:
  • \( \mathbf{u}_1 \cdot \mathbf{u}_2 = 0 \)
The result was indeed zero, confirming orthogonality. An appealing property of orthogonal sets is that they are easy to work with, especially in vector spaces. Each vector can be managed separately, allowing for more straightforward calculations.
Dot Product
The dot product, also known as the scalar product, is a fundamental operation involving two vectors. It provides a way to mathematically determine the extent to which two vectors align. This is calculated as
  • \(\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3 + \ldots \)
where \( u_i\) and \( v_i\) are the components of vectors \(\mathbf{u}\) and \(\mathbf{v}\), respectively. If the dot product is zero, as in our orthogonal vectors, it confirms they are perpendicular. This property is instrumental when checking whether vectors form an orthogonal set.
Projection Formula
The projection formula helps us find the "shadow" that one vector casts onto another in the same vector space. To project vector \(\mathbf{y}\) onto vector \(\mathbf{u}\), use:
  • \( \text{proj}_{\mathbf{u}}(\mathbf{y}) = \frac{\mathbf{y} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u} \)
This formula calculates the component of \(\mathbf{y}\) that aligns with \(\mathbf{u}\). For example, projecting \(\mathbf{y}\) onto \(\mathbf{u}_1\), we found:
  • \( \frac{7}{14} \mathbf{u}_1 = \left[\frac{3}{2}, -\frac{1}{2}, 1\right] \)
By projecting onto each vector in the orthogonal set, we can build the full projection of \(\mathbf{y}\) onto the span of those vectors.
Linear Combinations
In linear algebra, a linear combination consists of adding multiples of vectors together. This can form new vectors and represent vector spaces. For example, if you have vectors \(\mathbf{u}_1\) and \(\mathbf{u}_2\), a linear combination is
  • \( c_1\mathbf{u}_1 + c_2\mathbf{u}_2 \)
where \(c_1\) and \(c_2\) are scalars. In our context, these scalars come from the components found during the projection process.The total orthogonal projection:
  • \( \text{proj}_{\text{Span}\{\mathbf{u}_1, \mathbf{u}_2\}}(\mathbf{y}) = \text{proj}_{\mathbf{u}_1}(\mathbf{y}) + \text{proj}_{\mathbf{u}_2}(\mathbf{y}) \)
shows how working with linear combinations in the context of vector projections can construct solutions within vector spaces.

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Most popular questions from this chapter

In Exercises 17 and \(18, A\) is an \(m \times n\) matrix and \(b\) is in \(\mathbb{R}^{m} .\) Mark each statement True or False. Justify each answer. a. The general least-squares problem is to find an \(\mathbf{x}\) that makes \(A \mathbf{x}\) as close as possible to \(\mathbf{b}\) . b. A least-squares solution of \(A \mathbf{x}=\mathbf{b}\) is a vector \(\hat{\mathbf{x}}\) that satisfies \(A \hat{\mathbf{x}}=\hat{\mathbf{b}},\) where \(\hat{\mathbf{b}}\) is the orthogonal projection of \(\mathbf{b}\) onto \(\operatorname{Col} A\) c. A least-squares solution of \(A \mathbf{x}=\mathbf{b}\) is a vector \(\hat{\mathbf{x}}\) such that \(\|\mathbf{b}-A \mathbf{x}\| \leq\|\mathbf{b}-A \hat{\mathbf{x}}\|\) for all \(\mathbf{x}\) in \(\mathbb{R}^{n}\) . d. Any solution of \(A^{T} A \mathbf{x}=A^{T} \mathbf{b}\) is a least-squares solution of \(A \mathbf{x}=\mathbf{b}\) . e. If the columns of \(A\) are linearly independent, then the equation \(A \mathbf{x}=\mathbf{b}\) has exactly one least-squares solution.

Find the distance between \(\mathbf{u}=\left[\begin{array}{r}{0} \\ {-5} \\\ {2}\end{array}\right]\) and \(\mathbf{z}=\left[\begin{array}{r}{-4} \\ {-1} \\\ {8}\end{array}\right]\)

Find the distance between \(\mathbf{x}=\left[\begin{array}{c}{10} \\\ {-3}\end{array}\right]\) and \(\mathbf{y}=\left[\begin{array}{c}{-1} \\\ {-5}\end{array}\right]\)

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 \(A .\)

In Exercises \(1-4,\) find a least-squares solution of \(A \mathbf{x}=\mathbf{b}\) by (a) constructing the normal equations for \(\hat{\mathbf{x}}\) and (b) solving for \(\hat{\mathbf{x}}\) . $$ A=\left[\begin{array}{rr}{1} & {3} \\ {1} & {-1} \\ {1} & {1}\end{array}\right], \mathbf{b}=\left[\begin{array}{l}{5} \\ {1} \\\ {0}\end{array}\right] $$
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