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Chapter 7: Problem 43

Verify that if \(W\) is a subspace of an inner product space \(V\) and \(\mathbf{v}\) is in \(V\), then perp \(_{w}(\mathbf{v})\) is orthogonal to all \(\mathbf{w}\) in \(W\)

Short Answer

Expert verified
Yes, \(\mathbf{v}_{W^\perp}\) is orthogonal to all \\(\mathbf{w}\) in \\(W\\).

Step by step solution

01

Understand the Definitions

A subspace \(W\) of an inner product space \(V\) is a subset of \(V\) that is closed under addition and scalar multiplication. The perp \(W^\perp\) refers to the orthogonal complement of \(W\), which consists of vectors in \(V\) that are orthogonal to every vector in \(W\).
02

Express perp \(_{W} (\mathbf{v})\)

Let \(\mathbf{v}\) be a vector in \(V\). The component of \(\mathbf{v}\) orthogonal to \(W\), denoted as \(\mathbf{v}_{W^\perp}\), is defined such that \(\mathbf{v} = \mathbf{v}_W + \mathbf{v}_{W^\perp}\), where \(\mathbf{v}_W\) is in \(W\).
03

Use the Orthogonal Decomposition Property

According to the orthogonal decomposition theorem, each vector in \(V\) can be uniquely expressed as the sum of a vector in \(W\) and a vector in \(W^\perp\). Therefore, \(\mathbf{v}_{W^\perp}\) is orthogonal to every vector in \(W\).
04

Verify Orthogonality

To show orthogonality, take any vector \(\mathbf{w}\) in \(W\). The inner product \(\langle \mathbf{v}_{W^\perp}, \mathbf{w} \rangle = 0\) because \(\mathbf{v}_{W^\perp}\) is in \(W^\perp\), confirming that \(\mathbf{v}_{W^\perp}\) is orthogonal to \(\mathbf{w}\).
05

Conclusion

From the above proof, \(\mathbf{v}_{W^\perp}\) is orthogonal to any vector \(\mathbf{w}\) in \(W\) by definition of orthogonal complement. Thus, the statement is verified.

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Key Concepts

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

Inner Product Space
An inner product space is a vector space equipped with an additional structure called an inner product. This inner product is a special function that takes two vectors and returns a scalar. The inner product allows us to measure angles and lengths, which are not inherently part of vector spaces.
  • Bilinearity: The inner product is linear in each of its arguments separately.
  • Symmetric: Switching the order of the terms does not change the value, except on the imaginary part in complex spaces.
  • Positive-definiteness: The inner product of a vector with itself is always positive unless the vector is the zero vector.
The most common example of an inner product is the dot product in Euclidean space, which lets us compute the angle between two vectors and determine orthogonality.
Subspace
A subspace is a smaller vector space that is part of a larger vector space and inherits its operations. To be a subspace, a subset must follow certain rules, namely closure under addition and scalar multiplication.
  • Zero Vector: A subspace must include the zero vector.
  • Closed Under Addition: Adding any two vectors in the subspace should yield a vector still within the subspace.
  • Closed Under Scalar Multiplication: Multiplying any vector in the subspace by a scalar still results in a vector in the subspace.
In our example, the subspace, denoted as \(W\), serves as a reference from which orthogonal complements are defined.
Orthogonal Decomposition Theorem
The orthogonal decomposition theorem states that every vector in an inner product space can be uniquely decomposed into two components: one that lies within a given subspace \(W\), and another that lies in the orthogonal complement of \(W\), noted as \(W^\perp\).
  • Unique Decomposition: Every vector \(\mathbf{v}\) can be expressed as \(\mathbf{v} = \mathbf{v}_W + \mathbf{v}_{W^\perp}\), where \(\mathbf{v}_W\) is in \(W\) and \(\mathbf{v}_{W^\perp}\) is in \(W^\perp\).
  • Orthogonality: The vectors \(\mathbf{v}_{W}\) and \(\mathbf{v}_{W^\perp}\) are orthogonal. This means their inner product is zero, \(\langle \mathbf{v}_W, \mathbf{v}_{W^\perp} \rangle = 0\).
  • Practical Use: This theorem aids in simplifying complex vector problems into manageable parts by working within subspaces and their complements.
The theorem is fundamental in proving properties like orthogonality and is essential in mathematical areas such as linear algebra and functional analysis.

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Most popular questions from this chapter

Compute the pseudoinverse of \(A\) $$A=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]$$

In Exercises \(31-36,\langle\mathbf{u}, \mathbf{v}\rangle\) is an inner product. In Exercises \(31-\) \(34,\) prove that the given statement is an identity $$\langle\mathbf{u}+\mathbf{v}, \mathbf{u}-\mathbf{v}\rangle=\|\mathbf{u}\|^{2}-\|\mathbf{v}\|^{2}$$

Show that the least squares solution of \(A \mathbf{x}=\mathbf{b}\) is not unique and solve the normal equations to find all the least squares solutions. $$A=\left[\begin{array}{rrrr} 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 1 \\ 0 & -1 & 1 & 1 \\ 1 & -1 & 1 & 0 \end{array}\right], \mathbf{b}=\left[\begin{array}{r} 1 \\\ -3 \\ 2 \\ 4 \end{array}\right]$$

Show that if \(A=U \Sigma V^{T}\) is an SVD of \(A\), then the left singular vectors are eigenvectors of \(A A^{T}\).

Find the singular values of the given matrix. $$A=\left[\begin{array}{lll}2 & 0 & 1 \\\0 & 2 & 0\end{array}\right]$$
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