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Chapter 5: Problem 26

Let \(\left[\mathbf{v}_{1}, \ldots, \mathbf{v}_{n}\right\\}\) be an orthogonal basis for \(\mathbb{R}^{n}\) and let \(W=\operatorname{span}\left(\mathbf{v}_{1}, \ldots, \mathbf{v}_{k}\right) .\) Is it necessarily true that \(W^{\perp}=\) \(\operatorname{span}\left(\mathbf{v}_{\mathrm{k}+1}, \ldots, \mathbf{v}_{n}\right) ?\) Either prove that it is true or find a counterexample.

Short Answer

Expert verified
Yes, it is true that \(W^\perp = \operatorname{span}(\mathbf{v}_{k+1}, \ldots, \mathbf{v}_n)\) in this case.

Step by step solution

01

Understanding the Problem

We are given an orthogonal basis \(\{ \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \}\) for \(\mathbb{R}^n\) and we need to determine if the orthogonal complement \(W^\perp\) of a subspace \(W\) can be expressed as the span of remaining basis vectors \(\{ \mathbf{v}_{k+1}, \ldots, \mathbf{v}_n \}\).
02

Define Subspace and Orthogonal Complement

\(W\) is defined as the span of the first \(k\) vectors \(\{ \mathbf{v}_1, \ldots, \mathbf{v}_k \}\). We define \(W^\perp\) as the set of all vectors \(\mathbf{u} \) in \(\mathbb{R}^n\) such that \(\langle \mathbf{u}, \mathbf{w} \rangle = 0\) for every \(\mathbf{w} \in W\).
03

Orthogonality and Properties of Basis

Since \(\{ \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \}\) is an orthogonal basis, each vector is orthogonal to the others. Thus, any vector in the span of \(\{ \mathbf{v}_{k+1}, \ldots, \mathbf{v}_n \}\) is orthogonal to any vector in the span of \(\{ \mathbf{v}_1, \ldots, \mathbf{v}_k \}\).
04

Establish Orthogonal Complement Relation

Because any \(\mathbf{u} \in \operatorname{span}(\mathbf{v}_{k+1}, \ldots, \mathbf{v}_n)\) is orthogonal to every vector in \(W = \operatorname{span}(\mathbf{v}_1, \ldots, \mathbf{v}_k)\), we have \(\operatorname{span}(\mathbf{v}_{k+1}, \ldots, \mathbf{v}_n) \subseteq W^\perp\).
05

Completeness and Dimension Argument

Since \(\{ \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \}\) is a basis for \(\mathbb{R}^n\), the total dimension is \(n\). \(W\) has dimension \(k\), so \(W^\perp\) would naturally have dimension \(n-k\). \(\operatorname{span}(\mathbf{v}_{k+1}, \ldots, \mathbf{v}_n)\) also has the same dimension \(n-k\), as it is spanned by \(n-k\) vectors. This confirms \(W^\perp = \operatorname{span}(\mathbf{v}_{k+1}, \ldots, \mathbf{v}_n)\).

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Key Concepts

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

Orthogonal Basis
An orthogonal basis is a special type of basis in a vector space where each basis vector is orthogonal, or perpendicular, to all the others. In simpler terms, it means that the dot product between any two different vectors in this set is zero. This property allows for simplified calculations, especially when working with projections and decomposing vectors. If the basis vectors are not only orthogonal but also of unit length, the set is called an orthonormal basis.
  • Orthogonal basis vectors simplify computations because they remove cross-terms when calculating projections.
  • They help in easier visualization of geometrical concepts as they align to coordinate axes in a nice way.
  • An orthogonal basis makes it straightforward to express a vector as a linear combination of the basis vectors.
  • This concept is widely used in various applications like image processing and machine learning, where data is often represented in orthogonal coordinates.
Subspace
A subspace is a set of vectors that itself forms a vector space within a larger vector space. To be a subspace, this set of vectors must satisfy three conditions: it contains the zero vector, it is closed under addition, and it is closed under scalar multiplication. Subspaces are fundamental in linear algebra because they allow us to study smaller components or aspects of a larger vector space.
Here鈥檚 why understanding subspaces is important:
  • They help understand the structure of linear transformations, as transformations may map onto subspaces.
  • Subspaces can be used to analyze components of vectors, such as identifying patterns or simplifications.
  • Identifying a subspace helps in simplifying problems like linear equations and projections, making it easier to find solutions.
Linear Span
The linear span of a set of vectors is the set of all possible linear combinations that can be made with these vectors. If you imagine vectors as arrows, then the span is the shape that you could fill in using these arrows, pointing in different directions and at various lengths. It effectively captures the limits to which these vectors can stretch if combined in different ways.
Understanding the linear span is crucial because:
  • It defines the smallest subspace containing a given set of vectors, helping determine the reach or "span" of these vectors.
  • The span provides insights into the freedom of movement or directionality that a set of vectors has within a space.
  • Linear spans are used in solving linear equations, as they can represent the set of all possible solutions to a system.
  • Knowing a set's span aids in tasks like dimension calculation, orthogonal complements, and vector decompositions.
Dimension
The dimension of a vector space, in simple terms, is the number of vectors in its basis. This "basis" is essentially the smallest set of vectors that can be combined to form any vector in the space. Determining the dimension is akin to measuring the size or complexity of the vector space.
Here鈥檚 why dimension matters:
  • The dimension provides crucial information about the "degree of freedom" or independent directions available in a space - think of it as the number of axes in a coordinate system.
  • A higher dimension implies more potential movement directions and complexity, while a lower dimension indicates restrictions.
  • In the context of the orthogonal complement, knowing the dimensions of a subspace and its complement informs us about how they partition the entire space.
  • Understanding dimensions helps in various fields like data science, where high-dimensional spaces need reduction to make data computable.

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Most popular questions from this chapter

Sometimes the graph of a quadratic equation is a straight line, a pair of straight lines, or a single point. We refer to such a graph as a degenerate conic. It is also possible that the equation is not satisfied for any values of the variables, in which case there is no graph at all and we refer to the conic. as an imaginary conic. Identify the conic with the given equation as either degenerate or imaginary and, where possible, sketch the graph. $$3 x^{2}+y^{2}=0$$

Identify the graph of the given equation. $$2 x^{2}+y^{2}-8=0$$

Let \(W\) be a subspace of \(\mathbb{R}^{n}\) and \(\mathbf{v}\) a vector in \(\mathbb{R}^{n}\). Suppose that \(\mathbf{w}\) and \(\mathbf{w}^{\prime}\) are orthogonal vectors with \(\mathbf{w}\) in \(W\) and that \(\mathbf{v}=\mathbf{w}+\mathbf{w}^{\prime} .\) Is it necessarily true that \(\mathbf{w}^{\prime}\) is in \(W^{\perp}\) ? Either prove that it is true or find a counterexample.

Find a symmetric \(3 \times 3\) matrix with eigenvalues \(\lambda_{1}, \lambda_{2},\) and \(\lambda_{3}\) and corresponding orthogonal eigenvectors \(\mathbf{v}_{1}, \mathbf{v}_{2},\) and \(\mathbf{v}_{3}\) $$\begin{array}{l} \lambda_{1}=1, \lambda_{2}=2, \lambda_{3}=3, \mathbf{v}_{1}=\left[\begin{array}{l} 1 \\ 1 \\ 0 \end{array}\right], \mathbf{v}_{2}=\left[\begin{array}{r} 1 \\ -1 \\ 1 \end{array}\right] \\ \mathbf{v}_{3}=\left[\begin{array}{r} -1 \\ 1 \\ 2 \end{array}\right] \end{array}$$

Identify the quadric with the given equation and give its equation in standard form. $$-x^{2}-y^{2}-z^{2}+4 x y+4 x z+4 y z=12$$
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