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Chapter 2: Problem 2

The Pythagorean theorem asserts that for a set of \(n\) orthogonal vectors \(\left\\{x_{i}\right\\}\) \\[ \left\|\sum_{i=1}^{n} x_{i}\right\|^{2}=\sum_{i=1}^{n}\left\|x_{i}\right\|^{2} \\] (a) Prove this in the case \(n=2\) by an explicit computation of \(\left\|x_{1}+x_{2}\right\|^{2}\) (b) Show that this computation also establishes the general case, by induction.

Short Answer

Expert verified
For the case \(n=2\), use orthogonality to show that \[\left\|x_{1}+x_{2}\right\|^{2} = \left\|x_{1}\right\|^{2} + \left\|x_{2}\right\|^{2}\]. Then, use induction to generalize this to any \(n\).

Step by step solution

01

State the problem

Given the Pythagorean theorem for orthogonal vectors \(\left\{x_{i}\right\}\) we need to prove two things: (a) Prove the theorem for the case \(n=2\) by computing \(\left\|x_{1}+x_{2}\right\|^{2}\). (b) Use induction to establish the theorem for a general case.
02

Consider the case n=2

We start by computing \(\left\|x_{1}+x_{2}\right\|^{2}\). This is the first step to proving the theorem when \(n=2\).
03

Express and expand the norm

Express the norm squared as follows: \[\left\|x_{1}+x_{2}\right\|^{2} = (x_1 + x_2) \bullet (x_1 + x_2)\] Use the property of the dot product and expand this: \[(x_1 + x_2) \bullet (x_1 + x_2) = x_1 \bullet x_1 + 2x_1 \bullet x_2 + x_2 \bullet x_2\]
04

Apply orthogonality

Since the vectors are orthogonal, \((x_1 \bullet x_2) = 0\). Thus, \[\left\|x_{1}+x_{2}\right\|^{2} = x_1 \bullet x_1 + x_2 \bullet x_2\] This simplifies to \[\left\|x_{1}+x_{2}\right\|^{2} = \left\|x_{1}\right\|^{2} + \left\|x_{2}\right\|^{2}\]. Therefore, part (a) is proven.
05

Base case for induction

We have proven that for \(n=2\), \[\left\|\textbf{x}_{1}+\textbf{x}_{2}\right\|^{2} = \left\|\textbf{x}_{1}\right\|^{2} + \left\|\textbf{x}_{2}\right\|^{2}\]
06

Inductive hypothesis

Assume that the theorem holds for \(n=k\). In other words, assume \[\left\|\textbf{x}_{1}+\textbf{x}_{2}+...+\textbf{x}_{k}\right\|^{2} = \left\|\textbf{x}_{1}\right\|^{2} + \left\|\textbf{x}_{2}\right\|^{2} + ... + \left\|\textbf{x}_{k}\right\|^{2}\]
07

Inductive step

To prove it for \(n=k+1\), consider: \[\left\|\textbf{x}_{1}+\textbf{x}_{2}+...+\textbf{x}_{k}+\textbf{x}_{k+1}\right\|^{2} = (\textbf{x}_{1}+\textbf{x}_{2}+...+\textbf{x}_{k}) + \textbf{x}_{k+1}\right\|^{2}\] Use the previous results: \[= \left\|\left(\textbf{x}_{1}+\textbf{x}_{2}+...+\textbf{x}_{k}\right)\right|^{2} + \left\|\textbf{x}_{k+1}\right|^{2}\] Use the inductive hypothesis: \[= \left\|\textbf{x}_{1}\right|^{2} + \left\|\textbf{x}_{2}\right|^{2} + ... + \left\|\textbf{x}_{k}\right|^{2} + \left\|\textbf{x}_{k+1}\right|^{2}\]
08

Conclusion

By induction, the Pythagorean theorem holds for any \(n\). We have proven it for \(n=2\) and shown that if it holds for \(n=k\) then it holds for \(n=k+1\). Therefore, \[\left\| \sum_{i=1}^{n} \textbf{x}_{i}\right\|^{2} = \sum_{i=1}^{n} \left\| \textbf{x}_{i}\right\|^{2}\] for any \(n\).

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Key Concepts

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

Orthogonal Vectors
Understanding orthogonal vectors is crucial to grasp the Pythagorean theorem's application in vector spaces. Orthogonal vectors are vectors that are perpendicular to each other. In mathematical terms, two vectors \( \textbf{x}_1 \) and \( \textbf{x}_2 \) are orthogonal if their dot product is zero: \[ \textbf{x}_1 \cdot \textbf{x}_2 = 0 \] This means they form a right angle (90 degrees) with each other. In the context of the Pythagorean theorem, we use this property to simplify expressions and prove relationships.
Inductive Proof
An inductive proof is a method of proving statements for all natural numbers. It involves two key steps: the base case and the inductive step. First, we prove the statement for an initial value (usually n=1 or n=2). Then, we assume the statement is true for some arbitrary case n=k and use this assumption to prove it for the next case n=k+1. Together, these steps establish that the statement holds for all natural numbers. This method is widely used in mathematics because it allows us to build a proof incrementally.
Dot Product Expansion
The dot product, also known as the scalar product, is a way to multiply two vectors that results in a scalar. For vectors \( \textbf{a} = (a_1, a_2, ..., a_n) \) and \( \textbf{b} = (b_1, b_2, ..., b_n) \), their dot product is calculated as: \[ \textbf{a} \cdot \textbf{b} = a_1b_1 + a_2b_2 + ... + a_nb_n \] In our problem, we use the dot product to expand the norm squared of the sum of vectors: \[ \left\| \textbf{x}_1 + \textbf{x}_2 \right\|^{2} = (\textbf{x}_1 + \textbf{x}_2) \cdot (\textbf{x}_1 + \textbf{x}_2) \] By expanding this, we can simplify using the orthogonality property.
Base Case and Inductive Step
For our proof by induction, we start with the base case. We showed that the Pythagorean theorem holds for n=2 by directly computing and simplifying the norm squared. Next, we assume it holds for n=k. This is our inductive hypothesis. To complete our proof, we must prove it for n=k+1. By adding another orthogonal vector to our sum, we use the inductive hypothesis and the orthogonality property to extend the theorem to n=k+1, completing our proof.
Vector Norms
A vector norm gives a measure of the length or size of the vector. For a vector \( \textbf{v} = (v_1, v_2, ..., v_n) \), its norm is calculated as: \[ \left\| \textbf{v} \right\| = \sqrt {v_1^2 + v_2^2 + ... + v_n^2} \] In the context of the Pythagorean theorem for orthogonal vectors, the norm is used to show how the lengths of individual vectors relate to the length of their sum. The norm squared is particularly useful: \[ \left\| \textbf{v} \right\|^2 = v_1^2 + v_2^2 + ... + v_n^2 \] This quadratic form is the foundation of our computations and proofs.

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Most popular questions from this chapter

Let \(A \in \mathbb{C}^{m \times m}\) be hermitian. An eigenvector of \(A\) is a nonzero vector \(x \in \mathbb{C}^{m}\) such that \(A x=\lambda x\) for some \(\lambda \in \mathbb{C},\) the corresponding eigenvalue. (a) Prove that all eigenvalues of \(A\) are real (b) Prove that if \(x\) and \(y\) are eigenvectors corresponding to distinct eigenvalues then \(x\) and \(y\) arc orthogonal.

Show that if a matrix \(A\) is both triangular and unitary, then it is diagonal.

A Hadamard matrix is a matrix whose entries are all 卤1 and whose transpose is equal to its inverse times a constant factor. It is known that if \(A\) is a Hadamard matrix of dimension \(m>2,\) then \(m\) is a multiple of \(4 .\) It is not known, however, whether there is a Hadamard matrix for every such \(m\) though examples have been found for all cases \(m \leq 424\) Show that the following recursive description provides a Hadamard matrix of each dimension \(m=2^{k}, k=0,1,2, \ldots\) \\[ H_{0}=\left[\begin{array}{l} 1 \\ H_{k} \end{array}\right], \quad H_{k+1}=\left[\begin{array}{rr} H_{k} & H_{k} \\ -H_{k} \end{array}\right] \\]

What can be said about the eigenvalues of a unitary matrix?

If \(u\) and \(v\) are \(m\) -vectors, the matrix \(A=I+u v^{*}\) is known as a rank- one perturbation of the identity. Show that if \(A\) is nonsingular, then its inverse has the form \(A^{-1}=I+\alpha u v^{*}\) for some scalar \(\alpha,\) and give an expression for \(\alpha\) For what \(u\) and \(v\) is \(A\) singular? If it is singular, what is \(\operatorname{mull}(A) ?\)
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