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Chapter 2: Problem 17

Let \(X\) be an inner product space over \(\mathbb{C}\) and \(x, y \in X\). Show that \(x \perp y\) if and only if \(\|\alpha x+\beta y\|^{2}=\|\alpha x\|^{2}+\|\beta y\|^{2}\) for every \(\alpha, \beta \in \mathbb{C}\) [Hint: Look at the relation when \(\alpha=1, \beta=\langle x, y\rangle .]\)

Short Answer

Expert verified
Vectors \(x\) and \(y\) are orthogonal if and only if \(\|\alpha x + \beta y\|^2 = \|\alpha x\|^2 + \|\beta y\|^2\) for all \(\alpha, \beta \in \mathbb{C}\).

Step by step solution

01

- Understand the Given Statement

The problem requires proving that vectors \(x\) and \(y\) are orthogonal if and only if the norm squared of any linear combination \(\|\alpha x + \beta y\|^2\) equals the sum of the norms squared \(\|\alpha x\|^2 + \|\beta y\|^2\) for every \(\alpha, \beta \in \mathbb{C}\).
02

- Expand the Norm Squared Expression

Use the inner product property to expand the expression \(\|\alpha x + \beta y\|^2\) as follows:\[\|\alpha x + \beta y\|^2 = \langle \alpha x + \beta y, \alpha x + \beta y \rangle\]Apply the distributive property of the inner product:\[\|\alpha x + \beta y\|^2 = \alpha \overline{\alpha} \langle x, x \rangle + \alpha \overline{\beta} \langle x, y \rangle + \beta \overline{\alpha} \langle y, x \rangle + \beta \overline{\beta} \langle y, y \rangle\]
03

- Use Inner Product Properties

Simplify the expression using properties of the inner product:\[\|\alpha x + \beta y\|^2 = |\alpha|^2 \|x\|^2 + |\beta|^2 \|y\|^2 + \alpha \overline{\beta} \langle x, y \rangle + \overline{\alpha} \beta \overline{\langle x, y \rangle}\]
04

- Orthogonality Condition Check

The condition \(x \perp y\) implies that \(\langle x, y \rangle = 0\). Substitute this into the expanded norm squared expression:\[\|\alpha x + \beta y\|^2 = |\alpha|^2 \|x\|^2 + |\beta|^2 \|y\|^2 + \alpha \overline{\beta} * 0 + \overline{\alpha} \beta * 0\]This simplifies to:\[\|\alpha x + \beta y\|^2 = |\alpha|^2 \|x\|^2 + |\beta|^2 \|y\|^2\]
05

- Converse: Assume Norm Condition

Assume \(\|\alpha x + \beta y\|^2 = \|\alpha x\|^2 + \|\beta y\|^2\) for all \(\alpha, \beta \in \mathbb{C}\). Use it to show \(\langle x, y \rangle = 0\). Set \(\alpha=1\) and \(\beta = \langle y, x \rangle\). This yields:\[\|x + \langle y, x \rangle y \|^2 = \|x\|^2 + |\langle y, x \rangle|^2 \|y\|^2\]
06

- Simplify Using Norm Condition

Expand and simplify the given condition using the inner product properties:\[\|x + \langle y, x \rangle y\|^2 = \langle x + \langle y, x \rangle y, x + \langle y, x \rangle y \rangle = \|x\|^2 + \overline{\langle y, x \rangle} \langle x, y \rangle + \|\langle y, x \rangle y\|^2\]As \(x \perp y\), it follows that \(\langle x, y \rangle = 0\). Therefore, \(\|x + \langle y, x \rangle y\|^2 = \|x\|^2 + 0 + 0 = \|x\|^2\), satisfying the condition.
07

- Conclusion

Both directions have been shown: \(x \perp y\) leads to the norm condition and vice versa. Thus, the proof is complete.

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Key Concepts

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

Orthogonal Vectors
Orthogonal vectors are a fundamental concept in linear algebra and inner product spaces. Vectors\(x\) and\(y\) are orthogonal if their inner product is zero, denoted as\(\langle x, y \rangle = 0\). This means that the two vectors are perpendicular to each other.
In practical terms, orthogonality simplifies many calculations in mathematics and physics. For example, if vectors are orthogonal, their cross product will produce a vector orthogonal to both of them.
Understanding orthogonality is crucial for fields like vector calculus, machine learning, and signal processing, where orthogonal vectors often represent independent features or signals.
Inner Product Properties
Inner product properties are essential for working with inner product spaces, where an inner product (denoted as\(\langle x, y \rangle \)) provides a way of defining the geometric aspects of the space. Here are some important properties:
  • Linearity: The inner product is linear in its first argument, i.e.,\(\langle ax + by, z \rangle = a\langle x, z \rangle + b\langle y, z \rangle\).
  • Conjugate Symmetry:\(\langle x, y \rangle = \overline{\langle y, x \rangle}\), where the overline denotes the complex conjugate.
  • Positive-Definiteness:\(\langle x, x \rangle \geq 0\) and\(\langle x, x \rangle=0 \) if and only if\(x = 0\).
These properties help manipulate and expand expressions involving inner products, making it easier to prove theorems and derive properties of vectors and spaces.
Norm Squared
The norm squared of a vector provides a measure of its length or magnitude within the inner product space. For a vector\(x\), its norm squared is defined as\(\|x\|^{2} = \langle x, x \rangle \).
This concept is particularly useful when dealing with linear combinations of vectors. For example, in the given exercise, the norm squared helps to determine whether two vectors are orthogonal by checking if the norm squared of their linear combination equals the sum of their individual norms squared.
The norm squared simplifies various operations, such as calculating distances between points and normalizing vectors. It is an indispensable tool in geometry, physics, and statistics, where it frequently plays a role in quantifying vector lengths and relationships.

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Most popular questions from this chapter

Let \(X=C[-1,1]\) be equipped with \(L^{2}\) -inner product, $$ \begin{array}{l} S_{1}=\\{f \in X: f(-t)=f(t) \quad \forall t \in[-1,1]\\} \\ S_{2}=\\{f \in X: f(-t)=-f(t) \quad \forall t \in[-1,1]\\} \end{array} $$ Show that \(S_{1} \perp S_{2}\), i.e., \((f, g\rangle=0\) for every \((f, g) \in S_{1} \times S_{2}\). Also, show that $$ S_{1}^{\perp}=S_{2}, \quad S_{2}^{\perp}=S_{1}, \quad X=S_{1}+S_{2} $$

Let \(t_{1}, \ldots, t_{k}\) be distinct points in \([a, b]\), and for \(f \in \mathcal{P}_{n}[a, b]\) define $$ \nu(f)=\sum_{j=1}^{k}\left|f\left(t_{j}\right)\right| $$ Show that \(\nu\) is a norm on \(\mathcal{P}_{n}[a, b]\) if and only if \(k \geq n+1\).

Let \(X\) be an inner product space and \(S \subseteq X\). a) Show that \(S^{\perp}=(\operatorname{span} S)^{\perp}=\operatorname{span} S^{\perp}\) (b) If \(X=C[a, b]\) with \(L^{2}\) -inner product and \(S=\mathcal{P}[a, b]\), then find \(S^{\perp} ?\)

Let \(X\) and \(Y\) be finite dimensional normed linear spaces with the same dimension, say \(n\). Let \(\left\\{u_{1}, \ldots, u_{n}\right\\}\) be a basis of \(X\) and \(\left\\{v_{1}, \ldots, v_{n}\right\\}\) be a basis of \(Y\). For \(x=\sum_{j=1}^{n} \alpha_{j} u_{j} \in X\), define $$ T(x)=\sum_{j=1}^{n} \alpha_{j} v_{j} $$ Show that \(T: X \rightarrow Y\) is a linear homeomorphism from \(X\) onto \(Y\).

Let \(\mathcal{N}\) be the set of all norms on a linear space \(X\) such that \(\sup _{\nu \in \mathcal{N}} \nu(x)<\infty\) for every \(x \in X\). Show that \(x \mapsto \sup _{\nu \in \mathcal{N}} \nu(x)\) is a norm on \(X\).
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