Join IDNLearn.com and start getting the answers you've been searching for. Discover the reliable solutions you need with help from our comprehensive and accurate Q&A platform.
Sagot :
To verify that [tex]\( \| \cdot \|_2 \)[/tex] is a norm on [tex]\( \mathbb{C}^n \)[/tex], we need to check if it satisfies the following four properties for all [tex]\( \mathbf{z}, \mathbf{w} \in \mathbb{C}^n \)[/tex] and [tex]\( \alpha \in \mathbb{C} \)[/tex]:
1. Positive Definiteness: [tex]\( \| \mathbf{z} \|_2 \ge 0 \)[/tex] and [tex]\( \| \mathbf{z} \|_2 = 0 \)[/tex] if and only if [tex]\( \mathbf{z} = \mathbf{0} \)[/tex].
2. Homogeneity: [tex]\( \| \alpha \mathbf{z} \|_2 = |\alpha| \| \mathbf{z} \|_2 \)[/tex].
3. Triangle Inequality: [tex]\( \| \mathbf{z} + \mathbf{w} \|_2 \le \| \mathbf{z} \|_2 + \| \mathbf{w} \|_2 \)[/tex].
4. Symmetry: [tex]\( \| \mathbf{z} \|_2 = \| \mathbf{z} \|_2 \)[/tex] for all [tex]\( \mathbf{z} \in \mathbb{C}^n \)[/tex].
Let's verify each property step by step:
### 1. Positive Definiteness
Given [tex]\( \mathbf{z} = (z_1, z_2, \ldots, z_n) \in \mathbb{C}^n \)[/tex],
[tex]\[ \| \mathbf{z} \|_2 = \left( \sum_{i=1}^n |z_i|^2 \right)^{1/2}. \][/tex]
- Non-negativity: The term inside the sum, [tex]\( |z_i|^2 \)[/tex], is always non-negative (since [tex]\( |z_i| \)[/tex] is the modulus of a complex number and is non-negative). Thus, the sum of non-negative terms is also non-negative and taking the square root of a non-negative number gives another non-negative number.
[tex]\[ \| \mathbf{z} \|_2 \ge 0. \][/tex]
- Zero vector: If [tex]\( \mathbf{z} = \mathbf{0} \)[/tex], all components [tex]\( z_i = 0 \)[/tex], so
[tex]\[ \| \mathbf{0} \|_2 = \left( \sum_{i=1}^n |0|^2 \right)^{1/2} = \left( 0 \right)^{1/2} = 0. \][/tex]
- Conversely, if [tex]\( \| \mathbf{z} \|_2 = 0 \)[/tex], then
[tex]\[ \left( \sum_{i=1}^n |z_i|^2 \right)^{1/2} = 0. \][/tex]
This means that
[tex]\[ \sum_{i=1}^n |z_i|^2 = 0. \][/tex]
Since each [tex]\( |z_i|^2 \ge 0 \)[/tex] and a sum of non-negative terms is zero only if each term is zero, we have [tex]\( |z_i| = 0 \)[/tex] for all [tex]\( i \)[/tex]. Therefore, [tex]\( z_i = 0 \)[/tex] for all [tex]\( i \)[/tex], implying that [tex]\( \mathbf{z} = \mathbf{0} \)[/tex].
Thus, [tex]\( \| \mathbf{z} \|_2 \ge 0 \)[/tex] and [tex]\( \| \mathbf{z} \|_2 = 0 \)[/tex] if and only if [tex]\( \mathbf{z} = \mathbf{0} \)[/tex].
### 2. Homogeneity
Given [tex]\( \alpha \in \mathbb{C} \)[/tex] and [tex]\( \mathbf{z} = (z_1, z_2, \ldots, z_n) \in \mathbb{C}^n \)[/tex],
[tex]\[ \| \alpha \mathbf{z} \|_2 = \left( \sum_{i=1}^n |\alpha z_i|^2 \right)^{1/2}. \][/tex]
By the properties of the modulus,
[tex]\[ |\alpha z_i| = |\alpha| |z_i|, \][/tex]
so
[tex]\[ \| \alpha \mathbf{z} \|_2 = \left( \sum_{i=1}^n (|\alpha| |z_i|)^2 \right)^{1/2} = \left( |\alpha|^2 \sum_{i=1}^n |z_i|^2 \right)^{1/2} = |\alpha| \left( \sum_{i=1}^n |z_i|^2 \right)^{1/2} = |\alpha| \| \mathbf{z} \|_2. \][/tex]
This confirms the homogeneity property.
### 3. Triangle Inequality
Given [tex]\( \mathbf{z} = (z_1, z_2, \ldots, z_n) \)[/tex] and [tex]\( \mathbf{w} = (w_1, w_2, \ldots, w_n) \in \mathbb{C}^n \)[/tex],
[tex]\[ \| \mathbf{z} + \mathbf{w} \|_2 = \left( \sum_{i=1}^n |z_i + w_i|^2 \right)^{1/2}. \][/tex]
Using the Minkowski inequality (a generalization of the triangle inequality for complex numbers),
[tex]\[ |z_i + w_i|^2 \le (|z_i| + |w_i|)^2. \][/tex]
Therefore,
[tex]\[ \sum_{i=1}^n |z_i + w_i|^2 \le \sum_{i=1}^n (|z_i| + |w_i|)^2. \][/tex]
Expanding this sum,
[tex]\[ \sum_{i=1}^n (|z_i| + |w_i|)^2 = \sum_{i=1}^n (|z_i|^2 + 2|z_i||w_i| + |w_i|^2). \][/tex]
Thus,
[tex]\[ \sum_{i=1}^n |z_i + w_i|^2 \le \sum_{i=1}^n |z_i|^2 + 2 \sum_{i=1}^n |z_i||w_i| + \sum_{i=1}^n |w_i|^2. \][/tex]
By applying the Cauchy-Schwarz inequality,
[tex]\[ \sum_{i=1}^n |z_i||w_i| \le \left( \sum_{i=1}^n |z_i|^2 \right)^{1/2} \left( \sum_{i=1}^n |w_i|^2 \right)^{1/2}, \][/tex]
we get
[tex]\[ \sum_{i=1}^n |z_i + w_i|^2 \le \left( \sum_{i=1}^n |z_i|^2 \right) + 2 \left( \sum_{i=1}^n |z_i|^2 \right)^{1/2} \left( \sum_{i=1}^n |w_i|^2 \right)^{1/2} + \left( \sum_{i=1}^n |w_i|^2 \right). \][/tex]
Taking the square root of both sides, we get
[tex]\[ \| \mathbf{z} + \mathbf{w} \|_2 \le \| \mathbf{z} \|_2 + \| \mathbf{w} \|_2. \][/tex]
### 4. Symmetry
For any [tex]\( \mathbf{z} \in \mathbb{C}^n \)[/tex],
[tex]\[ \| \mathbf{z} \|_2 = \left( \sum_{i=1}^n |z_i|^2 \right)^{1/2}. \][/tex]
Clearly, this expression does not depend on any property that would change if [tex]\( \mathbf{z} \)[/tex] were ordered differently. Hence, [tex]\( \| \mathbf{z} \|_2 = \| \mathbf{z} \|_2 \)[/tex] is inherently true.
Therefore, [tex]\( \| \cdot \|_2 \)[/tex] satisfies all four properties of a norm, and thus, [tex]\( \| \cdot \|_2 \)[/tex] is indeed a norm on [tex]\( \mathbb{C}^n \)[/tex].
1. Positive Definiteness: [tex]\( \| \mathbf{z} \|_2 \ge 0 \)[/tex] and [tex]\( \| \mathbf{z} \|_2 = 0 \)[/tex] if and only if [tex]\( \mathbf{z} = \mathbf{0} \)[/tex].
2. Homogeneity: [tex]\( \| \alpha \mathbf{z} \|_2 = |\alpha| \| \mathbf{z} \|_2 \)[/tex].
3. Triangle Inequality: [tex]\( \| \mathbf{z} + \mathbf{w} \|_2 \le \| \mathbf{z} \|_2 + \| \mathbf{w} \|_2 \)[/tex].
4. Symmetry: [tex]\( \| \mathbf{z} \|_2 = \| \mathbf{z} \|_2 \)[/tex] for all [tex]\( \mathbf{z} \in \mathbb{C}^n \)[/tex].
Let's verify each property step by step:
### 1. Positive Definiteness
Given [tex]\( \mathbf{z} = (z_1, z_2, \ldots, z_n) \in \mathbb{C}^n \)[/tex],
[tex]\[ \| \mathbf{z} \|_2 = \left( \sum_{i=1}^n |z_i|^2 \right)^{1/2}. \][/tex]
- Non-negativity: The term inside the sum, [tex]\( |z_i|^2 \)[/tex], is always non-negative (since [tex]\( |z_i| \)[/tex] is the modulus of a complex number and is non-negative). Thus, the sum of non-negative terms is also non-negative and taking the square root of a non-negative number gives another non-negative number.
[tex]\[ \| \mathbf{z} \|_2 \ge 0. \][/tex]
- Zero vector: If [tex]\( \mathbf{z} = \mathbf{0} \)[/tex], all components [tex]\( z_i = 0 \)[/tex], so
[tex]\[ \| \mathbf{0} \|_2 = \left( \sum_{i=1}^n |0|^2 \right)^{1/2} = \left( 0 \right)^{1/2} = 0. \][/tex]
- Conversely, if [tex]\( \| \mathbf{z} \|_2 = 0 \)[/tex], then
[tex]\[ \left( \sum_{i=1}^n |z_i|^2 \right)^{1/2} = 0. \][/tex]
This means that
[tex]\[ \sum_{i=1}^n |z_i|^2 = 0. \][/tex]
Since each [tex]\( |z_i|^2 \ge 0 \)[/tex] and a sum of non-negative terms is zero only if each term is zero, we have [tex]\( |z_i| = 0 \)[/tex] for all [tex]\( i \)[/tex]. Therefore, [tex]\( z_i = 0 \)[/tex] for all [tex]\( i \)[/tex], implying that [tex]\( \mathbf{z} = \mathbf{0} \)[/tex].
Thus, [tex]\( \| \mathbf{z} \|_2 \ge 0 \)[/tex] and [tex]\( \| \mathbf{z} \|_2 = 0 \)[/tex] if and only if [tex]\( \mathbf{z} = \mathbf{0} \)[/tex].
### 2. Homogeneity
Given [tex]\( \alpha \in \mathbb{C} \)[/tex] and [tex]\( \mathbf{z} = (z_1, z_2, \ldots, z_n) \in \mathbb{C}^n \)[/tex],
[tex]\[ \| \alpha \mathbf{z} \|_2 = \left( \sum_{i=1}^n |\alpha z_i|^2 \right)^{1/2}. \][/tex]
By the properties of the modulus,
[tex]\[ |\alpha z_i| = |\alpha| |z_i|, \][/tex]
so
[tex]\[ \| \alpha \mathbf{z} \|_2 = \left( \sum_{i=1}^n (|\alpha| |z_i|)^2 \right)^{1/2} = \left( |\alpha|^2 \sum_{i=1}^n |z_i|^2 \right)^{1/2} = |\alpha| \left( \sum_{i=1}^n |z_i|^2 \right)^{1/2} = |\alpha| \| \mathbf{z} \|_2. \][/tex]
This confirms the homogeneity property.
### 3. Triangle Inequality
Given [tex]\( \mathbf{z} = (z_1, z_2, \ldots, z_n) \)[/tex] and [tex]\( \mathbf{w} = (w_1, w_2, \ldots, w_n) \in \mathbb{C}^n \)[/tex],
[tex]\[ \| \mathbf{z} + \mathbf{w} \|_2 = \left( \sum_{i=1}^n |z_i + w_i|^2 \right)^{1/2}. \][/tex]
Using the Minkowski inequality (a generalization of the triangle inequality for complex numbers),
[tex]\[ |z_i + w_i|^2 \le (|z_i| + |w_i|)^2. \][/tex]
Therefore,
[tex]\[ \sum_{i=1}^n |z_i + w_i|^2 \le \sum_{i=1}^n (|z_i| + |w_i|)^2. \][/tex]
Expanding this sum,
[tex]\[ \sum_{i=1}^n (|z_i| + |w_i|)^2 = \sum_{i=1}^n (|z_i|^2 + 2|z_i||w_i| + |w_i|^2). \][/tex]
Thus,
[tex]\[ \sum_{i=1}^n |z_i + w_i|^2 \le \sum_{i=1}^n |z_i|^2 + 2 \sum_{i=1}^n |z_i||w_i| + \sum_{i=1}^n |w_i|^2. \][/tex]
By applying the Cauchy-Schwarz inequality,
[tex]\[ \sum_{i=1}^n |z_i||w_i| \le \left( \sum_{i=1}^n |z_i|^2 \right)^{1/2} \left( \sum_{i=1}^n |w_i|^2 \right)^{1/2}, \][/tex]
we get
[tex]\[ \sum_{i=1}^n |z_i + w_i|^2 \le \left( \sum_{i=1}^n |z_i|^2 \right) + 2 \left( \sum_{i=1}^n |z_i|^2 \right)^{1/2} \left( \sum_{i=1}^n |w_i|^2 \right)^{1/2} + \left( \sum_{i=1}^n |w_i|^2 \right). \][/tex]
Taking the square root of both sides, we get
[tex]\[ \| \mathbf{z} + \mathbf{w} \|_2 \le \| \mathbf{z} \|_2 + \| \mathbf{w} \|_2. \][/tex]
### 4. Symmetry
For any [tex]\( \mathbf{z} \in \mathbb{C}^n \)[/tex],
[tex]\[ \| \mathbf{z} \|_2 = \left( \sum_{i=1}^n |z_i|^2 \right)^{1/2}. \][/tex]
Clearly, this expression does not depend on any property that would change if [tex]\( \mathbf{z} \)[/tex] were ordered differently. Hence, [tex]\( \| \mathbf{z} \|_2 = \| \mathbf{z} \|_2 \)[/tex] is inherently true.
Therefore, [tex]\( \| \cdot \|_2 \)[/tex] satisfies all four properties of a norm, and thus, [tex]\( \| \cdot \|_2 \)[/tex] is indeed a norm on [tex]\( \mathbb{C}^n \)[/tex].
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. IDNLearn.com has the solutions to your questions. Thanks for stopping by, and see you next time for more reliable information.
