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Chapter 6: Problem 1

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\. \(\|x\|_{\text {sup }}=\max (|-3|,|1|,|2|)=3\)

Short Answer

Expert verified
The Euclidean norm is \( \sqrt{14} \), the 1-norm is 6, and the infinity norm is 3.

Step by step solution

01

Calculate the Euclidean Norm

The Euclidean norm (or 2-norm) of a vector is calculated as the square root of the sum of the squares of its components. For the vector \( x = \left[ \begin{array}{c} -3 \ 1 \ 2 \end{array} \right] \), compute \( \|x\| \) as follows: \[ \|x\| = \sqrt{(-3)^2 + (1)^2 + (2)^2} = \sqrt{14} \].
02

Calculate the 1-Norm

The 1-norm of a vector is the sum of the absolute values of its components. For the vector \( x = \left[ \begin{array}{c} -3 \ 1 \ 2 \end{array} \right] \), calculate the 1-norm as follows: \[ \|x\|_1 = |-3| + |1| + |2| = 3 + 1 + 2 = 6 \].
03

Calculate the Infinity Norm

The infinity norm (also known as the supremum norm) of a vector is the maximum absolute value of its components. For the vector \( x = \left[ \begin{array}{c} -3 \ 1 \ 2 \end{array} \right] \), calculate the infinity norm as follows: \[ \|x\|_{\text{sup}} = \max (|-3|, |1|, |2|) = 3 \].

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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, often referred to as the 2-norm, is one of the most commonly used vector norms. It calculates the 'straight-line' distance from the origin to the point represented by the vector. To compute it, you take the square root of the sum of the squares of each component in the vector.

Consider the vector \( x = \left[ \begin{array}{c} -3 \ 1 \ 2 \end{array} \right] \). Here's how you find its Euclidean norm step-by-step:

  • Square each component: \((-3)^2, (1)^2, (2)^2\)
  • Add these squares together: 9 + 1 + 4 = 14
  • Take the square root of the sum: \(\sqrt{14} \approx 3.74\)
As a result, \(\|x\| = \sqrt{14} \approx 3.74\). This value gives you an idea of how far the point is from the origin in three-dimensional space.
1-norm
The 1-norm (also called the Manhattan norm or Taxicab norm) measures the 'distance' by summing the absolute values of each component in the vector. It's as if you were walking along city blocks, hence the nickname 'Manhattan norm.'

For the vector \( x = \left[ \begin{array}{c} -3 \ 1 \ 2 \end{array} \right] \), you would calculate the 1-norm as follows:

  • Take the absolute value of each component: \(|-3|, |1|, |2|\)
  • Add these values together: 3 + 1 + 2 = 6
So, \(\|x\|_1 = 6\). This sum represents the total 'distance' you'd cover if you moved along the orthogonal axes (like moving along streets and avenues in a grid layout).
Infinity norm
The infinity norm (often called the max norm or supremum norm) is a measure that looks at the largest absolute value among the vector's components. It signifies the maximum 'distance' you could cover along any single axis.

For the vector \( x = \left[ \begin{array}{c} -3 \ 1 \ 2 \end{array} \right] \), finding the infinity norm involves these steps:

  • Take the absolute value of each component: \(|-3|, |1|, |2|\)
  • Determine the maximum value among them: \(3\)
Hence, \(\|x\|_{\text{sup}} = 3\). This value indicates the farthest distance along any single axis, which in this case is the x-axis direction represented by -3.

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Most popular questions from this chapter

Find the cosine of the angle between \(x\) and \(y\) where \(x=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right] y=\left[\begin{array}{l}2 \\ 2 \\ 2\end{array}\right]\) \(\begin{array}{l}\text { 2. } x=\left[\begin{array}{c}1 \\ -1 \\ 1 \\\ -1\end{array}\right] & y=\left[\begin{array}{c}-1 \\ 0 \\ 1 \\\ 0\end{array}\right] \\ \text { 3. } x=\left[\begin{array}{c}2 \\ 1 \\\ -2\end{array}\right] & y=\left[\begin{array}{c}4 \\ 2 \\\ -4\end{array}\right]\end{array}\)

Given vectors \(x, y, z \in \mathbb{R}^{n}\) and the scalar \(a \in \mathbb{R}\), prove the following identities: 1\. \(x+y=y+x\). Is vector addition commutative?

Compute and plot \(x+y\) and \(x-y\) for each of the following cases: 1\. \(x=\left[\begin{array}{l}1 \\ 3 \\ 2\end{array}\right], y=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]\) \(x=\left[\begin{array}{c}-1 \\ 3 \\ -2\end{array}\right], y=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]\) 3\. \(x=\left[\begin{array}{c}1 \\ -3 \\ 2\end{array}\right], y=\left[\begin{array}{l}1 \\ 3 \\ 2\end{array}\right]\)

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.

(MATLAB) Write a MATLAB function that accepts any vector \(x \in R^{n}\) and computes its norm and direction cosines. Make \(x\) an input variable, and make \(\|x\|\) and \(u_{x}\) output variables.
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