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Chapter 1: Problem 3

Let \(V\) be an inner product space, \(u, v \in V, u, v \neq \overrightarrow{0}\). (a) Show that \(\langle u, v\rangle=\|u\| \cdot\|v\|\) if and only if \(u=a v\) for some \(a \in \mathbb{C}\). (b) Show that \(\|u+v\|=\|u\|+\|v\|\) if and only if \(u=a v\) for some \(a \geq 0\).

Short Answer

Expert verified
(a) Equality implies linear dependence; (b) Equality implies positive alignment.

Step by step solution

01

Understanding the Inner Product Equality

To show that \( \langle u, v\rangle = \|u\| \cdot \|v\| \) implies that \( u = a v \) for some \( a \in \mathbb{C} \), we start by using the definition of the Cauchy-Schwarz inequality. The inequality tells us that \( |\langle u, v \rangle| \leq \|u\| \cdot \|v\| \), with equality if and only if \( u \) and \( v \) are linearly dependent. Since we have equality here, it must be that \( u = a v \). Conversely, if \( u = a v \), then substituting into the inner product gives \( \langle u, v \rangle = \|a v\| \cdot \|v\| = |a| \|v\|^2 = \|a v\| \cdot \|v\| \). Therefore, \( \langle u, v \rangle = \|u\| \cdot \|v\| \).
02

Investigating Norm Equality Condition

Consider the condition \( \|u+v\| = \|u\| + \|v\| \). Using the triangle inequality, \( \|u+v\| \leq \|u\| + \|v\| \), with equality if and only if \( u \) and \( v \) are positively correlated (or aligned in direction or opposite direction without being negative-scale). If \( u = a v \) with \( a \geq 0 \), then \( u \) and \( v \) are aligned in the same direction, and thus, \( \|u+v\| = \|u\| + \|v\| \). Conversely, if \( \|u+v\| = \|u\| + \|v\| \), then \( u \) and \( v \) must be positively proportional, i.e., \( u = a v \) for some \( a \geq 0 \).
03

Summary of Results

For part (a), the equivalence of \( \langle u, v\rangle = \|u\| \cdot \|v\| \) to \( u = a v \) follows directly from the Cauchy-Schwarz inequality indicating linear dependence. For part (b), the equality \( \|u+v\| = \|u\| + \|v\| \) reduces to the requirement that \( u \) and \( v \) align in direction, expressed as \( u = a v \) where \( a \geq 0 \).

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Key Concepts

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

Cauchy-Schwarz Inequality
The Cauchy-Schwarz Inequality is fundamental in vector spaces and helps us relate the inner product of two vectors to the product of their norms. This principle tells us that, for any vectors \( u \) and \( v \), the absolute value of their inner product is always less than or equal to the product of their norms:\[ \left| \langle u, v \rangle \right| \leq \|u\| \cdot \|v\|. \]
The key part of this inequality is its condition for equality. It holds with equality if and only if the vectors are linearly dependent, meaning one vector is a scalar multiple of the other: \( u = a v \) for some scalar \( a \). This condition of equality plays a crucial role in the problem, because it connects the inner product directly with the notion of vector alignment. When the inner product equals the product of norms, we know the vectors are in the same or opposite directions.
  • This helps in identifying when two vectors are scalar multiples of each other.
  • This principle explains why if \( \langle u, v \rangle = \|u\| \cdot \|v\| \), it implies \( u = a v \).
Linear Dependence
Linear dependence is a concept from linear algebra that deals with the relationship between vectors. Vectors are said to be linearly dependent if there is a non-trivial combination of them that results in the zero vector. Simplifying, it means some vector in the group can be expressed as a combination of others.
For two vectors, \( u \) and \( v \), linear dependence means there exists a scalar \( a \) such that:\[ u = a v. \]This means \( u \) can be formed by multiplying \( v \) with \( a \). In the context of our exercise, understanding linear dependence directly helps understand why equality in the Cauchy-Schwarz Inequality leads us to conclude \( u = a v \).
  • Linear dependence provides the condition necessary for the equal norm and inner product relationship.
  • Recognizing dependent vectors helps in simplifying many vector space problems by reducing terms.
Triangle Inequality
The Triangle Inequality is a principle that gives us limits on the norms (or lengths) of vector sums. For any vectors \( u \) and \( v \) in an inner product space, the inequality is expressed as:\[ \| u + v \| \leq \|u\| + \|v\|. \]This concept is visual and stems from considering the sum of two sides of a triangle being greater than or equal to the third side. In the context of inner product spaces, the triangle inequality reflects the intuitive idea that the length of one side of a triangle (\( \|u + v\| \)) can't be longer than the sum of the other two.
  • This inequality becomes an equality if and only if \( u \) and \( v \) are collinear in the same direction.
  • If \( \|u + v\| = \|u\| + \|v\| \), then \( u = a v \) with \( a \geq 0 \), indicating they are aligned without opposition.
This understanding aligns perfectly with part (b) of the problem, demonstrating when \( u \) and \( v \) follow this notion of linear or directional alignment.

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Most popular questions from this chapter

Let \(\mathcal{D}=\left\\{(x, y) \in \mathbb{R}^{2} \mid x^{2}+y^{2}<1\right\\}\) denote the open unit disk in \(\mathbb{R}^{2}\). Let \(C(\mathcal{D})\) be the collection of all complex-valued continuous functions \(f\) : \(\mathcal{D} \rightarrow \mathbb{C}\) defined on \(\mathcal{D}\). For each \(f, g \in C(\mathcal{D})\) set $$ \langle f, g\rangle=\iint_{\mathcal{D}} f(x, y) \overline{g(x, y)} d x d y . $$ For each \(n \in \mathbb{N}\) we define $$ f_{n}(x, y)=(x+i y)^{n} . $$ (a) Prove that \(\langle f, g\rangle\) is an inner product on \(C(\mathcal{D})\). (b) Prove that the sequence of functions \(\left\\{f_{n}\right\\}_{n=0}^{\infty}\) is an infinite orthogonal system on \(C(\mathcal{D})\).

Let \(C^{1}[0,1]\) denote the space of continuous functions \(f:[0,1] \rightarrow \mathbb{C}\) with a continuous first derivative on \([0,1]\). Set $$ \langle f, g\rangle=f(0) \overline{g(0)}+f^{\prime}(0) \overline{g^{\prime}(0)}+f(1) \overline{g(1)} . $$ (a) Is this an inner product on the subspace \(\mathcal{P}_{2}=\operatorname{span}\left\\{1, x, x^{2}\right\\}\) ? (b) Is it an inner product on \(C^{1}[0,1]\) ?

Let \(C[-1,2]\) denote the space of continuous complex-valued functions \(f:[-1,2] \rightarrow \mathbb{C}\). Which of the following define an inner product on \(C[-1,2]\), and which do not? Explain. (a) \(\langle f, g\rangle=\int_{-1}^{2}|f(t)+g(t)| d t\) (b) \(\langle f, g\rangle=\int_{-1}^{2} f(t) \overline{g(t)} d t+f\left(-\frac{1}{2}\right) \overline{g\left(-\frac{1}{2}\right)}\) (c) \(\langle f, g\rangle=3 \int_{-1}^{2} f(t) \overline{g(t)} d t\) (d) \(\langle f, g\rangle=f(0) \overline{g(0)}+f(1) \overline{g(1)}\)

Let \(C^{1}[0,1]\) be the space of continuous functions \(f:[0,1] \rightarrow \mathbb{C}\) with a continuous first derivative on \([0,1]\). (a) Prove that $$ \langle f, g\rangle=f(0) \cdot \overline{g(0)}+\int_{0}^{1} f^{\prime}(x) \overline{g^{\prime}(x)} d x $$ is an inner product on \(C^{1}[0,1]\). (b) Find an orthonormal system \(\left\\{h_{1}, h_{2}, h_{3}\right\\}\) in \(C^{1}[0,1]\), with respect to this inner product, for which $$ \operatorname{span}\left\\{h_{1}, h_{2}, h_{3}\right\\}=\operatorname{span}\left\\{1, x, x^{2}\right\\} . $$

Prove that the set $$ V=\\{f \mid f: \mathbb{R} \rightarrow \mathbb{R}, f \text { is absolutely integrable over } \mathbb{R}\\} $$ is a linear space over \(\mathbb{R}\).
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