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Chapter 6: Problem 4

Show that the Euclidean norm is the same as the p-norm with \(\mathrm{p}=2\) and that the 1 -norm is the p-norm with \(\mathrm{p}=1 .\) (It can also be shown that the sup-norm is like a p-norm with \(\mathrm{p}=\infty .\) )

Short Answer

Expert verified
The Euclidean norm is the p-norm with \(p=2\). The 1-norm is the p-norm with \(p=1\). The sup-norm is the p-norm with \(p=\text{∞}\).

Step by step solution

01

Understand the Euclidean Norm

The Euclidean norm (or 2-norm) of a vector \(\textbf{v} = [v_1, v_2, ..., v_n]\) is given by \(\textbf{v}_2 = \sqrt{v_1^2 + v_2^2 + ... + v_n^2}\).
02

Define the p-norm for p=2

The general p-norm of a vector \(\textbf{v} = [v_1, v_2, ..., v_n]\) is defined as \(\textbf{v}_p = (|v_1|^p + |v_2|^p + ... + |v_n|^p)^{1/p}\). For \(\textbf{p} = 2\), this becomes \(\textbf{v}_2 = (|v_1|^2 + |v_2|^2 + ... + |v_n|^2)^{1/2}\), which matches the definition of the Euclidean norm.
03

Explain the 1-norm

The 1-norm of a vector \(\textbf{v} = [v_1, v_2, ..., v_n]\) is given by \(\textbf{v}_1 = |v_1| + |v_2| + ... + |v_n|\).
04

Define the p-norm for p=1

For \(\textbf{p} = 1\), the p-norm becomes \(\textbf{v}_1 = (|v_1|^1 + |v_2|^1 + ... + |v_n|^1)^{1/1} = |v_1| + |v_2| + ... + |v_n|\), which matches the definition of the 1-norm.
05

Introduction to the sup-norm

The sup-norm (or infinity norm) of a vector \(\textbf{v} = [v_1, v_2, ..., v_n]\) is given by \(\textbf{v}_\text{∞} = \text{max}(|v_1|, |v_2|, ..., |v_n|)\).
06

Demonstrate the p-norm for p=∞

For \(\textbf{p} = \text{∞}\), the p-norm is defined as \(\textbf{v}_\text{∞} = \text{max}( |v_1|, |v_2|, ..., |v_n|) = (\text{lim}_{p \to \text{∞}} (|v_1|^p + |v_2|^p + ... + |v_n|^p)^{1/p} )\), which matches the definition of the sup-norm.

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Key Concepts

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

Euclidean Norm
The Euclidean Norm, also known as the 2-norm, is one of the most commonly used vector norms. It measures the 'length' of a vector in Euclidean space.
To calculate the Euclidean norm of a vector \(\textbf{v} = [v_1, v_2, ..., v_n]\), you use the formula: \[ \textbf{v}_2 = \sqrt{v_1^2 + v_2^2 + ... + v_n^2} \]
Essentially, it involves squaring each component of the vector, summing these squares, and then taking the square root of the result. This method is useful for computing distances in multi-dimensional spaces, making it a crucial concept in both mathematics and applied fields like physics and computer science.
p-norm
The p-norm is a general form that includes various norms, including the 1-norm and the Euclidean norm.
A vector \(\textbf{v} = [v_1, v_2, ..., v_n]\) can have many different p-norms, defined by: \[ \textbf{v}_p = (|v_1|^p + |v_2|^p + ... + |v_n|^p)^{1/p} \] Given different values of \(p\), you obtain different types of norms:
  • For \(p = 1\), you get the 1-norm.
  • For \(p = 2\), you get the Euclidean norm or 2-norm.
  • For \(p \to \text{∞}\), you get the infinity norm.
By changing \(p\), you can adapt the concept of distance to various contexts, making the p-norm a versatile mathematical tool.
1-norm
The 1-norm (also called the Manhattan norm or Taxicab norm) is a way to measure distance by summing the absolute values of the vector components. Given a vector \(\textbf{v} = [v_1, v_2, ..., v_n]\), the 1-norm is calculated as: \[ \textbf{v}_1 = |v_1| + |v_2| + ... + |v_n| \] This norm gets its name from the way distances are measured in a grid-like path layout, resembling the street grid of Manhattan.
Unlike the Euclidean norm, the 1-norm does not involve squaring or taking the square root of the values, making it computationally simpler in many instances. It is often used in optimization problems and linear programming.
Infinity Norm
The infinity norm, also known as the sup-norm or maximum norm, measures the largest absolute value among the components of a vector. For a vector \(\textbf{v} = [v_1, v_2, ..., v_n]\), the infinity norm is defined as: \[ \textbf{v}_\text{∞} = \text{max}(|v_1|, |v_2|, ..., |v_n|) \] This norm is useful in contexts where the largest element determines the magnitude of the vector. It is often used in fields like data analysis and error estimation where the largest deviation is of most interest.
Conceptually, it can be thought of as the \(\text{lim}_{p \to \text{∞}}\) of the p-norm.
Sup-Norm
The sup-norm is another term for the infinity norm. Both terms are interchangeable and mean the same thing. When discussing the sup-norm, you are essentially looking at the 'supremum' (or maximum) value among the absolute values of vector components.
For a vector \(\textbf{v} = [v_1, v_2, ..., v_n]\), the sup-norm is given by: \[ \textbf{v}_\text{∞} = \text{max}(|v_1|, |v_2|, ..., |v_n|) \] This norm is particularly useful in analyzing the worst-case scenario, making it crucial for various applications in optimization and numerical analysis. Like the infinity norm, it focuses on the largest element in the vector, providing a different perspective on measuring vector magnitude.

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Most popular questions from this chapter

Compute the norm and the direction cosines for the vector \(x=\left[\begin{array}{l}4 \\ 2 \\ 6\end{array}\right]\).

Compute and plot the scalar product ax when \(x=\left[\begin{array}{c}1 \\ 1 / 2 \\ l / 4\end{array}\right]\) for each of the following scalars: 1\. \(a=1\) 2\. \(a=-1\) 3\. \(a=-1 / 4\) 4\. \(a=2\)

Which of the following pairs of vectors is orthogonal: 1\. \(x=\left[\begin{array}{l}1 \\ 0 \\ 1\end{array}\right], y=\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]\) 2\. \(x=\left[\begin{array}{l}1 \\ 1 \\ 0\end{array}\right], y=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]\) 3\. \(x=e_{1}, y=e_{3}\) 4\. \(x=\left[\begin{array}{l}a \\ b\end{array}\right], y=\left[\begin{array}{c}-b \\ a\end{array}\right]\)

Consider the vector \(x=\left[\begin{array}{c}-3 \\ 1 \\\ 2\end{array}\right]\). Then 1\. ||\(x||=\left[(-3)^{2}+(1)^{2}+(2)^{2}\right]^{1 / 2}=(14)^{1 / 2}\) 2\. \(\|x\|_{1}=(|-3|+|1|+|2|)=6\) 3\. \(\|x\|_{\text {sup }}=\max (|-3|,|1|,|2|)=3\)

Use the Cauchy-Schwarz inequality to prove the triangle inequality, which states $$ \|x+y\| \leq\|x\|+\|y\| $$ Explain why this is called the triangle inequality.
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