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Chapter 10: Problem 11

Let \(\mathbf{v}\) and \(\mathbf{w}\) be vectors in an inner product space \(V .\) Show that: a. \(\mathbf{v}\) is orthogonal to \(\mathbf{w}\) if and only if \(\|\mathbf{v}+\mathbf{w}\|=\|\mathbf{v}-\mathbf{w}\|\) b. \(\mathbf{v}+\mathbf{w}\) and \(\mathbf{v}-\mathbf{w}\) are orthogonal if and only if \(\|\mathbf{v}\|=\|\mathbf{w}\| .\)

Short Answer

Expert verified
(a) \(\mathbf{v}\) and \(\mathbf{w}\) are orthogonal iff \(\|\mathbf{v}+\mathbf{w}\|=\|\mathbf{v}-\mathbf{w}\|\). (b) \(\|\mathbf{v}\|=\|\mathbf{w}\|\) iff \(\mathbf{v}+\mathbf{w}\) and \(\mathbf{v}-\mathbf{w}\) are orthogonal.

Step by step solution

01

Understand Orthogonality

Two vectors \(\mathbf{u}\) and \(\mathbf{v}\) are orthogonal if their inner product is zero, i.e., \(\langle \mathbf{u}, \mathbf{v} \rangle = 0\). This will be used as the key property for solving parts (a) and (b).
02

Prove Part (a) - "If" Direction

Assume \(\mathbf{v}\) is orthogonal to \(\mathbf{w}\). Then, \(\langle \mathbf{v}, \mathbf{w} \rangle = 0\). Calculate the norms: \[\begin{align*}\|\mathbf{v} + \mathbf{w}\|^2 &= \langle \mathbf{v} + \mathbf{w}, \mathbf{v} + \mathbf{w} \rangle = \langle \mathbf{v}, \mathbf{v} \rangle + \langle \mathbf{v}, \mathbf{w} \rangle + \langle \mathbf{w}, \mathbf{v} \rangle + \langle \mathbf{w}, \mathbf{w} \rangle \ &= \|\mathbf{v}\|^2 + 0 + 0 + \|\mathbf{w}\|^2 = \|\mathbf{v}\|^2 + \|\mathbf{w}\|^2.\end{align*}\] Similarly, \(\|\mathbf{v}-\mathbf{w}\|^2 = \|\mathbf{v}\|^2 + \|\mathbf{w}\|^2\), proving that \(\|\mathbf{v}+\mathbf{w}\| = \|\mathbf{v}-\mathbf{w}\|\).
03

Prove Part (a) - "Only If" Direction

Assume \(\|\mathbf{v}+\mathbf{w}\| = \|\mathbf{v}-\mathbf{w}\|\). Then: \[ \begin{align*}\|\mathbf{v}+\mathbf{w}\|^2 &= \langle \mathbf{v}+\mathbf{w}, \mathbf{v}+\mathbf{w} \rangle = \|\mathbf{v}\|^2 + 2\langle \mathbf{v}, \mathbf{w} \rangle + \|\mathbf{w}\|^2, \\|\mathbf{v}-\mathbf{w}\|^2 &= \langle \mathbf{v}-\mathbf{w}, \mathbf{v}-\mathbf{w} \rangle = \|\mathbf{v}\|^2 - 2\langle \mathbf{v}, \mathbf{w} \rangle + \|\mathbf{w}\|^2.\end{align*} \] Setting them equal: \[\|\mathbf{v}\|^2 + 2\langle \mathbf{v}, \mathbf{w} \rangle + \|\mathbf{w}\|^2 = \|\mathbf{v}\|^2 - 2\langle \mathbf{v}, \mathbf{w} \rangle + \|\mathbf{w}\|^2\], resulting in \(\langle \mathbf{v}, \mathbf{w} \rangle = 0\), which implies \(\mathbf{v}\) is orthogonal to \(\mathbf{w}\).
04

Prove Part (b) - "If" Direction

Assume \(\|\mathbf{v}\| = \|\mathbf{w}\|\). Calculate the inner product of \(\mathbf{v}+\mathbf{w}\) and \(\mathbf{v}-\mathbf{w}\): \[ \langle \mathbf{v}+\mathbf{w}, \mathbf{v}-\mathbf{w} \rangle = \langle \mathbf{v}, \mathbf{v} \rangle - \langle \mathbf{v}, \mathbf{w} \rangle + \langle \mathbf{w}, \mathbf{v} \rangle - \langle \mathbf{w}, \mathbf{w} \rangle. \]Since \(\|\mathbf{v}\| = \|\mathbf{w}\|\), these terms equal each other, thus: \[\langle \mathbf{v}+\mathbf{w}, \mathbf{v}-\mathbf{w} \rangle = \|\mathbf{v}\|^2 - \|\mathbf{w}\|^2 = 0\].
05

Prove Part (b) - "Only If" Direction

Assume \(\mathbf{v}+\mathbf{w}\) and \(\mathbf{v}-\mathbf{w}\) are orthogonal. This implies: \[ \langle \mathbf{v}+\mathbf{w}, \mathbf{v}-\mathbf{w} \rangle = \|\mathbf{v}\|^2 - \|\mathbf{w}\|^2 = 0\].Thus, \(\|\mathbf{v}\| = \|\mathbf{w}\|\).
06

Conclusion

We have shown that (a) \(\mathbf{v}\) is orthogonal to \(\mathbf{w}\) if and only if \(\|\mathbf{v}+\mathbf{w}\| = \|\mathbf{v}-\mathbf{w}\|\), and (b) \(\mathbf{v}+\mathbf{w}\) and \(\mathbf{v}-\mathbf{w}\) are orthogonal if and only if \(\|\mathbf{v}\| = \|\mathbf{w}\|\).

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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." The inner product is a special operation that allows for the measurement of angles and lengths within the space. Think of it as a way to "multiply" vectors to get a scalar, which helps in seeing how vectors relate to each other in terms of direction and magnitude.
  • The result of the inner product of two vectors is a real number (or a complex number in more advanced cases).
  • In mathematical terms, for vectors \( \mathbf{u} \) and \( \mathbf{v} \), their inner product is denoted as \( \langle \mathbf{u}, \mathbf{v} \rangle \).
  • This operation must satisfy specific properties: it must be commutative, linear, and positive-definite.

These properties ensure that the inner product is useful for defining concepts like angles between vectors and orthogonality, which means vectors are "perpendicular" in this context. The inner product helps determine when two vectors have zero interactions, in other words, when they are orthogonal (their inner product equals zero). This concept is foundational in proving problems related to orthogonality and norm equality.
Norm Equality
The norm of a vector is essentially its "length" or "size" within the vector space. When we discuss norm equality, we're talking about situations where the lengths of two vectors are identical, which can have important implications.
  • The norm is derived from the inner product, specifically as \( \| \mathbf{v} \| = \sqrt{\langle \mathbf{v}, \mathbf{v} \rangle} \).
  • Norm equality (\( \| \mathbf{v} \| = \| \mathbf{w} \| \)) suggests that vectors \( \mathbf{v} \) and \( \mathbf{w} \) are of the same magnitude, regardless of their directions.

In the context of the exercise, if vectors \( \mathbf{v} + \mathbf{w} \) and \( \mathbf{v} - \mathbf{w} \) are orthogonal, their norms being equal (\( \| \mathbf{v} \| = \| \mathbf{w} \| \)) helps in deducing certain geometric properties within the space. This equality is critical in proving part (b) of the exercise, where the orthogonality of sum and difference of vectors implies equal norms.
Orthogonal Vectors
Orthogonal vectors are vectors that meet at a right angle in a geometrical sense, or more technically, have an inner product of zero. When two vectors are orthogonal, they essentially "ignore" each other, as one has no projection onto the other. This is a fundamental concept in vector spaces, especially in understanding vector relationships.
  • When \( \langle \mathbf{v}, \mathbf{w} \rangle = 0 \), the vectors \( \mathbf{v} \) and \( \mathbf{w} \) are said to be orthogonal.
  • Orthogonality is not just about perpendicularity; it's a measure of "independence" between vectors, which is a cornerstone of linear algebra.

The exercise illustrates two important implications of orthogonality:
1. If \( \mathbf{v} \) is orthogonal to \( \mathbf{w} \), then it reflects in their norms (\( \|\mathbf{v} + \mathbf{w}\| = \|\mathbf{v} - \mathbf{w}\| \)).
2. Furthermore, when sum and difference vectors (\( \mathbf{v} + \mathbf{w} \) and \( \mathbf{v} - \mathbf{w} \)) are orthogonal, it indicates equal magnitudes (\( \|\mathbf{v}\| = \|\mathbf{w}\| \)). These points are key in advancing through the proof and understanding deeper mathematical structures involving orthogonal projections and decompositions.

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Most popular questions from this chapter

Using the inner product given by \(\langle p, q\rangle=\int_{0}^{1} p(x) q(x) d x\) on \(\mathbf{P}_{2},\) write \(\mathbf{v}\) as the sum of a vector in \(U\) and a vector in \(U^{\perp}\). a. \(\mathbf{v}=x^{2}, U=\operatorname{span}\\{x+1,9 x-5\\}\) b. \(\mathbf{v}=x^{2}+1, U=\operatorname{span}\\{1,2 x-1\\}\)

Let \(\mathbf{n} \neq \mathbf{0}\) and \(\mathbf{w} \neq \mathbf{0}\) be nonparallel vectors in \(\mathbb{R}^{3}\) (as in Chapter 4). a. Show that \(\left\\{\mathbf{n}, \mathbf{n} \times \mathbf{w}, \mathbf{w}-\frac{\mathbf{n} \cdot \mathbf{w}}{\|\mathbf{n}\|^{2}} \mathbf{n}\right\\}\) is an orthogo- nal basis of \(\mathbb{R}^{3}\). b. Show that \(\operatorname{span}\left\\{\mathbf{n} \times \mathbf{w}, \mathbf{w}-\frac{\mathbf{n} \cdot \mathbf{w}}{\|\mathbf{n}\|^{2}} \mathbf{n}\right\\}\) is the plane through the origin with normal \(\mathbf{n}\).

In each case find \(U^{\perp}\) and compute \(\operatorname{dim} U\) and \(\operatorname{dim} U^{\perp}\) a. \(U=\operatorname{span}\\{(1,1,2,0),(3,-1,2,1),\) (1,-3,-2,1)\\} in \(\mathbb{R}^{4}\) b. \(U=\operatorname{span}\\{(1,1,0,0)\\}\) in \(\mathbb{R}^{4}\) c. \(U=\operatorname{span}\\{1, x\\}\) in \(\mathbf{P}_{2}\) with \(\langle p, q\rangle=p(0) q(0)+p(1) q(1)+p(2) q(2)\) d. \(U=\operatorname{span}\\{x\\}\) in \(\mathbf{P}_{2}\) with \(\langle p, q\rangle=\int_{0}^{1} p(x) q(x) d x\) e. \(U=\operatorname{span}\left\\{\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right],\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right]\right\\}\) in \(\mathbf{M}_{22}\) with \(\langle X, Y\rangle=\operatorname{tr}\left(X Y^{T}\right)\) f. \(U=\operatorname{span}\left\\{\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}1 & 0 \\ 1 & 0\end{array}\right],\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]\right\\}\) in \(\mathbf{M}_{22}\) with \(\langle X, Y\rangle=\operatorname{tr}\left(X Y^{T}\right)\)

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 \(U^{\perp}=W\).

If \(T: V \rightarrow V\) is an isomorphism of the inner product space \(V\), show that $$ \langle\mathbf{v}, \mathbf{w}\rangle_{1}=\langle T(\mathbf{v}), T(\mathbf{w})\rangle $$ defines a new inner product \(\langle,\rangle_{1}\) on \(V\).
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