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Chapter 5: Problem 16

In Exercises \(15-18,\) find the orthogonal projection of v onto the subspace \(W\) spanned by the vectors \(\mathbf{u}_{i} .\) ( You may assume that the vectors \(\mathbf{u}_{i}\) are orthogonal. $$\mathbf{v}=\left[\begin{array}{r} 3 \\ 1 \\ -2 \end{array}\right], \mathbf{u}_{1}=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right], \mathbf{u}_{2}=\left[\begin{array}{r} 1 \\ -1 \\ 0 \end{array}\right]$$

Short Answer

Expert verified
The orthogonal projection is \( \begin{bmatrix} \frac{5}{3} \\ -\frac{1}{3} \\ \frac{2}{3} \end{bmatrix} \).

Step by step solution

01

Verify Orthogonality

Check if the vectors \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \) are orthogonal by calculating their dot product. Compute \( \mathbf{u}_1 \cdot \mathbf{u}_2 = (1)(1) + (1)(-1) + (1)(0) = 1 - 1 + 0 = 0 \). Since the dot product is zero, \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \) are orthogonal.
02

Calculate Projection onto \( \mathbf{u}_1 \)

To find the projection of \( \mathbf{v} \) onto \( \mathbf{u}_1 \), use the formula: \[ \text{proj}_{\mathbf{u}_1} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{u}_1}{\mathbf{u}_1 \cdot \mathbf{u}_1} \mathbf{u}_1 \]Calculate \( \mathbf{v} \cdot \mathbf{u}_1 = (3)(1) + (1)(1) + (-2)(1) = 3 + 1 - 2 = 2 \).Calculate \( \mathbf{u}_1 \cdot \mathbf{u}_1 = (1)(1) + (1)(1) + (1)(1) = 3 \).Thus, \( \text{proj}_{\mathbf{u}_1} \mathbf{v} = \frac{2}{3} \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} = \begin{bmatrix} \frac{2}{3} \ \frac{2}{3} \ \frac{2}{3} \end{bmatrix} \).
03

Calculate Projection onto \( \mathbf{u}_2 \)

To find the projection of \( \mathbf{v} \) onto \( \mathbf{u}_2 \), use the formula: \[ \text{proj}_{\mathbf{u}_2} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{u}_2}{\mathbf{u}_2 \cdot \mathbf{u}_2} \mathbf{u}_2 \]Calculate \( \mathbf{v} \cdot \mathbf{u}_2 = (3)(1) + (1)(-1) + (-2)(0) = 3 - 1 + 0 = 2 \).Calculate \( \mathbf{u}_2 \cdot \mathbf{u}_2 = (1)(1) + (-1)(-1) + (0)(0) = 1 + 1 = 2 \).Thus, \( \text{proj}_{\mathbf{u}_2} \mathbf{v} = \frac{2}{2} \begin{bmatrix} 1 \ -1 \ 0 \end{bmatrix} = \begin{bmatrix} 1 \ -1 \ 0 \end{bmatrix} \).
04

Calculate Total Orthogonal Projection

The orthogonal projection of \( \mathbf{v} \) onto the subspace \( W \) is the sum of the projections onto \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \).Thus, \[ \text{proj}_W \mathbf{v} = \begin{bmatrix} \frac{2}{3} \ \frac{2}{3} \ \frac{2}{3} \end{bmatrix} + \begin{bmatrix} 1 \ -1 \ 0 \end{bmatrix} = \begin{bmatrix} \frac{2}{3} + 1 \ \frac{2}{3} - 1 \ \frac{2}{3} + 0 \end{bmatrix} = \begin{bmatrix} \frac{5}{3} \ -\frac{1}{3} \ \frac{2}{3} \end{bmatrix} \].

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Key Concepts

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

Dot Product
When you're working with vectors, one of the first operations you'll encounter is the dot product. The dot product is a way to multiply two vectors together to get a single number, not a vector. This operation is crucial for determining many things about vectors, including their orthogonality.

For two vectors, say \( \mathbf{a} = [a_1, a_2, a_3] \) and \( \mathbf{b} = [b_1, b_2, b_3] \), the dot product is computed as follows:
  • Multiply the corresponding components: \( a_1b_1, a_2b_2, a_3b_3 \).
  • Sum the results: \( a_1b_1 + a_2b_2 + a_3b_3 \).
If the dot product equals zero, it means the vectors are orthogonal, which means they are perpendicular to each other. This property of the dot product makes it an excellent tool for understanding vector relationships.
Orthogonality
A key concept that comes up frequently in vector mathematics is orthogonality. Two vectors are orthogonal if their dot product is zero. Imagine two lines meeting at a right angle; this is similar to what orthogonality means for vectors.

Orthogonality is an important concept because it helps us understand vector space relationships.
  • If vectors are orthogonal, they won't affect each other in terms of projection, meaning projecting one onto the other results in no change or just zero.
  • In a subspace spanned by orthogonal vectors, any vector can be broken down into components along these orthogonal directions.
Understanding orthogonality can simplify many mathematical operations, especially when dealing with projections or constructing bases for vector spaces.
Vector Projection
Vector projection is used to "cast" a vector onto another. This concept is very much like casting a shadow; the shadow represents the direction and magnitude of the projection. By projecting one vector onto another, you find out how much of one vector goes in the direction of another vector.

To project a vector \(\mathbf{v}\) onto another vector \(\mathbf{u}\), you use the formula: \[\text{proj}_{\mathbf{u}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u} \]Key steps include:
  • Calculate the dot product \( \mathbf{v} \cdot \mathbf{u} \).
  • Divide by \( \mathbf{u} \cdot \mathbf{u} \), the dot product of \( \mathbf{u} \) with itself, to adjust the magnitude.
  • Multiply this scalar by \( \mathbf{u} \) to find the projection.
This process creates a vector pointing in the same direction as \(\mathbf{u}\) but has the length determined by how much \(\mathbf{v}\) "aligns" with \(\mathbf{u}\).
Subspace
In linear algebra, when we talk about subspaces, we're referring to a smaller space within a larger vector space. This smaller space, or subspace, is made up of all linear combinations of a given set of vectors.

Some properties of subspaces include:
  • They must pass through the origin, meaning a zero vector must be in the subspace.
  • For any vectors \(\mathbf{a}\) and \(\mathbf{b}\) in the subspace, any linear combination, such as \(c\mathbf{a} + d\mathbf{b}\) (where \(c\) and \(d\) are scalars), also resides in the subspace.
A classic example is the span of a set of vectors. If vectors are orthogonal, they can form an orthogonal basis for this subspace, simplifying many computations like projections.
Linear Combination
A linear combination involves combining several vectors using multiplication by scalars (numbers) and addition. In a sense, you can think of it as mixing vectors, just like mixing colors, to form new vectors.

For vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\), a linear combination looks like this: \[ c_1\mathbf{a} + c_2\mathbf{b} + c_3\mathbf{c} \] where \(c_1, c_2, \) and \(c_3\) are scalars. Concepts to Remember:
  • Any vector within a vector space or subspace can be composed as a linear combination of the basis vectors of that space.
  • The coefficients \( c_1, c_2, \) and \( c_3 \) determine how much one vector contributes to the resulting vector.
  • Linear combinations help us solve systems of equations and are fundamental to spanning and defining vector spaces.
By understanding linear combinations, you can efficiently navigate through vector spaces by constructing the vectors you need for your analysis.

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Most popular questions from this chapter

Find the symmetric matrix \(A\) associated with the given quadratic form. \(2 x^{2}-3 y^{2}+z^{2}-4 x z\)

Let \(S=\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{k}\right\\}\) be an orthonormal set in \(\mathbb{R}^{n},\) and let \(\mathbf{x}\) be a vector in \(\mathbb{R}^{n}\) (a) Prove that \\[ \|\mathbf{x}\|^{2} \geq\left|\mathbf{x} \cdot \mathbf{v}_{1}\right|^{2}+\left|\mathbf{x} \cdot \mathbf{v}_{2}\right|^{2}+\cdots+\left|\mathbf{x} \cdot \mathbf{v}_{k}\right|^{2} \\] (This inequality is called Bessel's Inequality.) (b) Prove that Bessel's Inequality is an equality if and only if \(\mathbf{x}\) is in \(\operatorname{span}(S)\)

Classify each of the quadratic forms as positive definite, positive semidefinite, negative definite negative semidefinite, or indefinite. $$x_{1}^{2}+x_{2}^{2}-x_{3}^{2}+4 x_{1} x_{2}$$

Classify each of the quadratic forms as positive definite, positive semidefinite, negative definite, negative semidefinite, or indefinite \(x_{1}^{2}+2 x_{2}^{2}\)

In Exercises \(11-14\), let \(W\) be the subspace spanned by the given vectors. Find a basis for \(W^{\perp}\) $$\mathbf{w}_{1}=\left[\begin{array}{r} 4 \\ 6 \\ -1 \\ 1 \\ -1 \end{array}\right], \mathbf{w}_{2}=\left[\begin{array}{r} 1 \\ 2 \\ 0 \\ 1 \\ -3 \end{array}\right], \mathbf{w}_{3}=\left[\begin{array}{r} 2 \\ 2 \\ 2 \\ -1 \\ 2 \end{array}\right]$$
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