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Chapter 10: Problem 12

Let \(U\) and \(W\) be subspaces of an \(n\) dimensional inner product space \(V .\) Suppose \(\langle\mathbf{u}, \mathbf{v}\rangle=0\) for all \(\mathbf{u} \in U\) and \(\mathbf{w} \in W\) and \(\operatorname{dim} U+\operatorname{dim} W=n\). Show that \(U^{\perp}=W\).

Short Answer

Expert verified
Orthogonality and dimension sum imply \(U^\perp = W\).

Step by step solution

01

Understanding Perpendicular Subspaces

Two subspaces of the inner product space, \(U\) and \(W\), are said to be perpendicular \((U \perp W)\) if every vector \(\mathbf{u} \in U\) is orthogonal to every vector \(\mathbf{w} \in W\). The problem states that such orthogonality exists if for all \(\mathbf{u} \in U\) and \(\mathbf{w} \in W\), \(\langle \mathbf{u}, \mathbf{w} \rangle = 0\).
02

Dimensions and Orthogonal Complements

Consider \(\operatorname{dim} U + \operatorname{dim}W = n\). In an \(n\)-dimensional space, the dimension of a subspace and its orthogonal complement add up to \(n\). Therefore, \(\operatorname{dim} U + \operatorname{dim} U^\perp = n\). Similarly, for any subspace \(W\), \(\operatorname{dim} W + \operatorname{dim} W^\perp = n\).
03

Concluding Equality of Orthogonal Complements

Given that \(\langle \mathbf{u}, \mathbf{w} \rangle = 0\) for all vectors in \(U\) and \(W\), any vector that is orthogonal to every vector in \(U\) must be in \(W\). Since \(\operatorname{dim} U + \operatorname{dim} W = n\), the dimension relationships imply that \(U^\perp\) must coincide with \(W\). Thus, \(U^\perp = W\).

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Key Concepts

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

Inner Product Space
In mathematics, an inner product space is a vector space with an additional structure called an inner product. This inner product is a way to multiply vectors together to get a scalar value. It provides a geometric interpretation to vector spaces and allows the definition of angles and lengths.
  • **Definition**: In an inner product space, each pair of vectors \(\mathbf{u}\) and \(\mathbf{v}\) is associated with a scalar \(\langle \mathbf{u}, \mathbf{v} \rangle\).
  • **Properties**: The inner product is typically designed to satisfy linearity in its first argument, conjugate symmetry, and positive-definiteness.
Understanding these spaces is crucial because they allow us to discuss concepts like orthogonality and projection in a precise way. When working inside an inner product space, concepts like orthogonal complements are immediately accessible, expanding our toolkit for analyzing vector subspaces.
Subspaces
A subspace is a specific kind of set in vector spaces, sticking to specific rules to make it part of the larger vector space.
  • **Subset Relationship**: Every subspace of a vector space \(V\) is essentially a non-empty subset of \(V\) that is itself a vector space under the operations of addition and scalar multiplication defined for \(V\).
  • **Closure Properties**: For any vectors \(\mathbf{u}, \mathbf{v}\) in a subspace \(U\), if you take \(c\) as any scalar, both \(\mathbf{u} + \mathbf{v}\) and \(c\mathbf{u}\) must also be in \(U\).
The key role of subspaces in mathematics emerges because they allow us to study smaller, more manageable sections of a larger vector space, making complex problems easier to solve by breaking them down.
Orthogonality
Orthogonality in vector spaces is a concept that refers to the perpendicularity of vectors. Two vectors are called orthogonal if their inner product equals zero.
  • **Perpendicular Vectors**: Orthogonal vectors, such as \(\mathbf{u}\) and \(\mathbf{v}\), satisfy the condition \(\langle \mathbf{u}, \mathbf{v} \rangle = 0\).
  • **Implications**: This relationship simplifies many mathematical processes, particularly when dealing with projections and decompositions of vectors.
  • **Applications**: Orthogonality is widely used in solving systems of equations and in algorithms like the Gram-Schmidt process, which generates an orthogonal basis from a set of linearly independent vectors.
Understanding orthogonality helps in visualizing geometric concepts in mathematics, providing clarity in many complex problems.
Dimension
Dimension is a fundamental concept related to the size or space a vector set can span.
  • **Definition**: The dimension of a vector space is the number of vectors in any basis for the space.
  • **Significance**: In an \(n\)-dimensional space, the dimension indicates how complex or diverse a space is regarding its span.
  • **Subspaces and Complements**: For every subspace \(U\) in \(V\), its orthogonal complement's dimension \(\operatorname{dim} U^\perp\) relates directly to the whole space: \(\operatorname{dim} U + \operatorname{dim} U^\perp = \operatorname{dim} V\).
Grasping the idea of dimensions aids in determining the structure and limits of vector spaces, which is crucial for solving linear algebra problems.
Vector Spaces
Vector spaces are a cornerstone concept in mathematics. They consist of a collection of vectors that can be added together and multiplied by numbers, called scalars.
  • **Structure**: A vector space must satisfy several axioms, like associativity, distributivity, and existence of an additive identity and inverses.
  • **Components**: Examples of vector spaces include the set of all real numbers, ordered pairs (like points in a plane), or functions meeting certain conditions.
  • **Significance**: They provide a universal language to describe and solve problems in numerous fields like computer science, physics, and engineering.
Studying vector spaces is essential as they form the mathematical framework allowing for a structured approach to solve equations and model real-world phenomena.

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Most popular questions from this chapter

In each case, determine which of axioms \(\mathrm{P} 1-\mathrm{P} 5\) fail to hold. a. \(V=\mathbb{R}^{2},\left\langle\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)\right\rangle=x_{1} y_{1} x_{2} y_{2}\) b. \(V=\mathbb{R}^{3}\) \(\left\langle\left(x_{1}, x_{2}, x_{3}\right),\left(y_{1}, y_{2}, y_{3}\right)\right\rangle=x_{1} y_{1}-x_{2} y_{2}+x_{3} y_{3}\) c. \(V=\mathbb{C},\langle z, w\rangle=z \bar{w}\), where \(\bar{w}\) is complex conjugation d. \(V=\mathbf{P}_{3},\langle p(x), q(x)\rangle=p(1) q(1)\) e. \(V=\mathbf{M}_{22},\langle A, B\rangle=\operatorname{det}(A B)\) f. \(V=\mathbf{F}[0,1],\langle f, g\rangle=f(1) g(0)+f(0) g(1)\)

Let \(\mathbf{D}_{n}\) denote the space of all functions from the set \(\\{1,2,3, \ldots, n\\}\) to \(\mathbb{R}\) with pointwise addition and scalar multiplication (see Exercise 6.3 .35 ). Show that \(\langle,\rangle\) is an inner product on \(\mathbf{D}_{n}\) if \(\langle\mathbf{f}, \mathbf{g}\rangle=f(1) g(1)+f(2) g(2)+\cdots+f(n) g(n)\)

Exercise \(\mathbf{1 0 . 1 . 1 3}\) In each case, find a symmetric matrix \(A\) such that \(\langle\mathbf{v}, \mathbf{w}\rangle=\mathbf{v}^{T} A \mathbf{w}\) $$ \begin{array}{l} \text { a. }\left\langle\left[\begin{array}{l} v_{1} \\ v_{2} \end{array}\right],\left[\begin{array}{c} w_{1} \\ w_{2} \end{array}\right]\right\rangle=v_{1} w_{1}+2 v_{1} w_{2}+2 v_{2} w_{1}+5 v_{2} w_{2} \\ \text { b. }\left\langle\left[\begin{array}{l} v_{1} \\ v_{2} \end{array}\right],\left[\begin{array}{c} w_{1} \\ w_{2} \end{array}\right]\right\rangle=v_{1} w_{1}-v_{1} w_{2}-v_{2} w_{1}+2 v_{2} w_{2} \end{array} $$ $$ \begin{array}{l} \text { c. }\left\langle\left[\begin{array}{l} v_{1} \\ v_{2} \\ v_{3} \end{array}\right],\left[\begin{array}{l} w_{1} \\ w_{2} \\ w_{3} \end{array}\right]\right\rangle=\begin{aligned} & 2 v_{1} w_{1}+v_{2} w_{2}+v_{3} w_{3}-v_{1} w_{2} \\ &-v_{2} w_{1}+v_{2} w_{3}+v_{3} w_{2} \\ \text { d. }\left\langle\left[\begin{array}{l} v_{1} \\ v_{2} \\ v_{3} \end{array}\right],\left[\begin{array}{l} w_{1} \\ w_{2} \\ w_{3} \end{array}\right]\right\rangle=& v_{1} w_{1}+2 v_{2} w_{2}+5 v_{3} w_{3} \\ &-2 v_{1} w_{3}-2 v_{3} w_{1} \end{aligned} \end{array} $$

Let \(v\) denote a vector in an inner product space \(V\). a. Show that \(W=\\{\mathbf{w} \mid \mathbf{w}\) in \(V,\langle\mathbf{v}, \mathbf{w}=0\\}\) is a subspace of \(V\). b. Let \(W\) be as in (a). If \(V=\mathbb{R}^{3}\) with the dot product, and if \(\mathbf{v}=(1,-1,2),\) find a basis for \(W\).

If \(A\) is a \(2 \times n\) matrix, let \(\mathbf{u}\) and \(\mathbf{v}\) denote the rows of \(A\). a. Show that \(A A^{T}=\left[\begin{array}{cc}\|\mathbf{u}\|^{2} & \mathbf{u} \cdot \mathbf{v} \\ \mathbf{u} \cdot \mathbf{v} & \|\mathbf{v}\|^{2}\end{array}\right]\). b. Show that \(\operatorname{det}\left(A A^{T}\right) \geq 0\).
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