Answer :
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].
