Math Problem Statement

Given linear independence and normalized vectors properties to prove conditions on dot products and spans for vectors u, v, w in R^n.

Solution

Let's go through the exercises and their requirements step-by-step.


Exercise 1

  1. Given: u,v,wRn\mathbf{u}, \mathbf{v}, \mathbf{w} \in \mathbb{R}^n are linearly independent, and ARn×nA \in \mathbb{R}^{n \times n} is an invertible matrix.

    • Objective: Show that {Au,Av,Aw}\{A\mathbf{u}, A\mathbf{v}, A\mathbf{w}\} are linearly independent.

    Solution Approach:

    • Start by recalling that since AA is invertible, it preserves linear independence.
    • Use the property that for an invertible matrix AA, if {u,v,w}\{ \mathbf{u}, \mathbf{v}, \mathbf{w} \} are linearly independent, then {Au,Av,Aw}\{A\mathbf{u}, A\mathbf{v}, A\mathbf{w}\} will also be linearly independent.
  2. Given: u,v,wRn\mathbf{u}, \mathbf{v}, \mathbf{w} \in \mathbb{R}^n are linearly independent.

    • Objective: Show that span(u,v)span(u,v,w)\text{span}(\mathbf{u}, \mathbf{v}) \neq \text{span}(\mathbf{u}, \mathbf{v}, \mathbf{w}).

    Solution Approach:

    • Since u,v,w\mathbf{u}, \mathbf{v}, \mathbf{w} are linearly independent, w\mathbf{w} does not lie in the span of {u,v}\{\mathbf{u}, \mathbf{v}\}.
    • Therefore, span(u,v,w)\text{span}(\mathbf{u}, \mathbf{v}, \mathbf{w}) is a three-dimensional space, while span(u,v)\text{span}(\mathbf{u}, \mathbf{v}) is two-dimensional, making the spans unequal.

Exercise 2

Let u,v,wRn\mathbf{u}, \mathbf{v}, \mathbf{w} \in \mathbb{R}^n (with n3n \geq 3) be three normalized vectors.

  1. (a) (1pt) Show that uw1\mathbf{u} \cdot \mathbf{w} \geq -1.

    Solution Approach:

    • Since both u\mathbf{u} and w\mathbf{w} are normalized vectors, their dot product is bounded by the Cauchy-Schwarz inequality.
    • Thus, 1uw1-1 \leq \mathbf{u} \cdot \mathbf{w} \leq 1, so uw1\mathbf{u} \cdot \mathbf{w} \geq -1.
  2. (b) (3pt) Now suppose that uv=12\mathbf{u} \cdot \mathbf{v} = -\frac{1}{2} and uw=13\mathbf{u} \cdot \mathbf{w} = -\frac{1}{3}. Show that vw23\mathbf{v} \cdot \mathbf{w} \geq -\frac{2}{3}.

    Solution Approach:

    • Apply the concept of angle relationships or geometric constraints based on the given dot products.
    • Use the triangle inequality in terms of angles between these vectors to find an inequality that includes vw\mathbf{v} \cdot \mathbf{w} in relation to uv\mathbf{u} \cdot \mathbf{v} and uw\mathbf{u} \cdot \mathbf{w}.
  3. (c) (1pt) Additionally, suppose that ww=0\mathbf{w} \cdot \mathbf{w} = 0. Complete an admissible triple {u,v}=e1,w=ej\{\mathbf{u}, \mathbf{v}\} = \mathbf{e}_1, \mathbf{w} = \mathbf{e}_j satisfying all the relations.

    Solution Approach:

    • This part seems to involve finding vectors that form a basis while satisfying the given conditions.
    • Use standard basis vectors like e1\mathbf{e}_1, ej\mathbf{e}_j, etc., to represent an orthonormal system that adheres to the given dot product constraints.

Would you like detailed solutions for any part, or have any specific questions?


Follow-Up Questions

  1. What is the significance of an invertible matrix preserving linear independence in a vector space?
  2. Why does the span of three linearly independent vectors differ from that of two?
  3. How does the Cauchy-Schwarz inequality apply to normalized vectors?
  4. What is the impact of specific dot products in determining angles between vectors?
  5. How would changing one of the conditions affect the solution?

Tip

Always check if the matrix is invertible when dealing with transformations and linear independence; invertibility often preserves independence properties.

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Math Problem Analysis

Mathematical Concepts

Linear Independence
Dot Product
Span of Vectors
Vector Normalization

Formulas

Dot product: u ⋅ v = |u||v|cos(θ)
Span: span(u, v) is the set of all linear combinations of u and v
Cauchy-Schwarz Inequality: |u ⋅ v| ≤ |u||v|

Theorems

Linear Independence Theorem
Invertible Matrix Theorem
Cauchy-Schwarz Inequality

Suitable Grade Level

Undergraduate Level (University Mathematics)