Chapter 1: Problem 2
In \(\mathrm{R}^{n}\) show that (a) \(2\|\mathbf{x}\|^{2}+2\|\mathbf{y}\|^{2}=\|\mathbf{x}+\mathbf{y}\|^{2}+\|\mathbf{x}-\mathbf{y}\|^{2}\) (This is known as the parallelogram law.) (b) \(\|\mathbf{x}-\mathbf{y}\|\|\mathbf{x}+\mathbf{y}\| \leq\|\mathbf{x}\|^{2}+\|\mathbf{y}\|^{2}\) (c) \(4\langle\mathbf{x}, \mathbf{y}\rangle=\|\mathbf{x}+\mathbf{y}\|^{2}-\|\mathbf{x}-\mathbf{y}\|^{2}\) (This is called the polarization identity.)
Short Answer
Step by step solution
Understanding the Norm and Scalar Product
Prove the Parallelogram Law
Apply Cauchy-Schwarz Inequality
Prove the Polarization Identity
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parallelogram Law
In mathematical terms, it states that for any vectors \(\mathbf{x}\) and \(\mathbf{y}\), the sum of the squares of the lengths equals the sum of the squares of the diagonals:
- \[2\|\mathbf{x}\|^{2}+2\|\mathbf{y}\|^{2}=\|\mathbf{x}+\mathbf{y}\|^{2}+\|\mathbf{x}-\mathbf{y}\|^{2}\]
Cauchy-Schwarz Inequality
This inequality for vectors \(\mathbf{x}\) and \(\mathbf{y}\) in \(\mathbb{R}^n\) is represented as:
- \(\langle \mathbf{x}, \mathbf{y} \rangle \leq \|\mathbf{x}\| \|\mathbf{y}\|\)
This implies that vectors relatively maintain their alignment at any operation scale. In our exercise, it manifests as
- \(\|\mathbf{x}-\mathbf{y}\|\|\mathbf{x}+\mathbf{y}\|\leq\|\mathbf{x}\|^{2}+\|\mathbf{y}\|^{2}\)
Polarization Identity
For any vectors \(\mathbf{x}\) and \(\mathbf{y}\) in \(\mathbb{R}^n\), the identity is expressed as:
- \[4\langle\mathbf{x}, \mathbf{y}\rangle=\|\mathbf{x}+\mathbf{y}\|^{2}-\|\mathbf{x}-\mathbf{y}\|^{2}\]
This transformation from angles and lengths provides deeper insights into symmetry and analytic geometry properties that these vectors possess.
Scalar Product
Defined as:
- \(\langle \mathbf{x}, \mathbf{y} \rangle = \sum_{i=1}^{n} x_i y_i\)
I understand concepts like angle cosine between vectors using this scalar translate effortlessly into diverse applications, ranging from physics to computer graphics.
Euclidean Norm
Its formula in vector terms is:
- \(\|\mathbf{v}\| = \sqrt{\langle\mathbf{v}, \mathbf{v}\rangle}\)
Distance and lengths defined by such norms allow precise calculations crucial for vectors and matrices, forming building blocks in fields such as image processing, optimization, and machine learning.