If x = (2, 1 + i, i) and y = (2 - i, 2, 1 + 2i) then verify Schwarz's inequality

Schwarz's inequality (Cauchy-Schwarz inequality) for vectors states that for any vectors \( \mathbf{x} \) and \( \mathbf{y} \) in a complex inner product space, the following inequality holds:

\[
| \langle \mathbf{x}, \mathbf{y} \rangle |^2 \leq \langle \mathbf{x}, \mathbf{x} \rangle \langle \mathbf{y}, \mathbf{y} \rangle
\]

Where:
- \( \langle \mathbf{x}, \mathbf{y} \rangle \) is the inner product of \( \mathbf{x} \) and \( \mathbf{y} \).
- \( | \langle \mathbf{x}, \mathbf{y} \rangle | \) is the magnitude of the inner product.

Given vectors:
\[
\mathbf{x} = (2, 1 + i, i), \quad \mathbf{y} = (2 - i, 2, 1 + 2i)
\]

### Step 1: Compute \( \langle \mathbf{x}, \mathbf{y} \rangle \)
The inner product of two vectors \( \mathbf{x} \) and \( \mathbf{y} \) in a complex space is:

\[
\langle \mathbf{x}, \mathbf{y} \rangle = \sum_{k=1}^{n} x_k \overline{y_k}
\]

Where \( \overline{y_k} \) is the complex conjugate of \( y_k \).

1. \( x_1 = 2 \), \( y_1 = 2 - i \):
   \[
   x_1 \overline{y_1} = 2(2 + i) = 4 + 2i
   \]

2. \( x_2 = 1 + i \), \( y_2 = 2 \):
   \[
   x_2 \overline{y_2} = (1 + i)(2) = 2 + 2i
   \]

3. \( x_3 = i \), \( y_3 = 1 + 2i \):
   \[
   x_3 \overline{y_3} = i(1 - 2i) = i - 2i^2 = i + 2 = 2 + i
   \]

Now, sum these values:
\[
\langle \mathbf{x}, \mathbf{y} \rangle = (4 + 2i) + (2 + 2i) + (2 + i) = 8 + 5i
\]

### Step 2: Compute \( | \langle \mathbf{x}, \mathbf{y} \rangle |^2 \)
First, find the magnitude of \( \langle \mathbf{x}, \mathbf{y} \rangle \):

\[
| \langle \mathbf{x}, \mathbf{y} \rangle | = |8 + 5i| = \sqrt{8^2 + 5^2} = \sqrt{64 + 25} = \sqrt{89}
\]

Now square the magnitude:
\[
| \langle \mathbf{x}, \mathbf{y} \rangle |^2 = 89
\]

### Step 3: Compute \( \langle \mathbf{x}, \mathbf{x} \rangle \) and \( \langle \mathbf{y}, \mathbf{y} \rangle \)

1. \( \langle \mathbf{x}, \mathbf{x} \rangle \):
   \[
   \langle \mathbf{x}, \mathbf{x} \rangle = (2)(\overline{2}) + (1 + i)(\overline{1 + i}) + (i)(\overline{i})
   \]
   \[
   = 2^2 + (1 + i)(1 - i) + i(-i) = 4 + (1^2 - i^2) + 1 = 4 + (1 + 1) + 1 = 4 + 2 + 1 = 7
   \]

2. \( \langle \mathbf{y}, \mathbf{y} \rangle \):
   \[
   \langle \mathbf{y}, \mathbf{y} \rangle = (2 - i)(\overline{2 - i}) + 2(\overline{2}) + (1 + 2i)(\overline{1 + 2i})
   \]
   \[
   = (2^2 + (-i)(i)) + 2^2 + (1^2 + 2^2) = (4 + 1) + 4 + (1 + 4) = 5 + 4 + 5 = 14
   \]

### Step 4: Verify the inequality
Now check if Schwarz's inequality holds:
\[
| \langle \mathbf{x}, \mathbf{y} \rangle |^2 \leq \langle \mathbf{x}, \mathbf{x} \rangle \langle \mathbf{y}, \mathbf{y} \rangle
\]

Substitute the values:
\[
89 \leq 7 \times 14 = 98
\]

Since \( 89 \leq 98 \), the Cauchy-Schwarz inequality is verified.

Would you like more details or have any questions?

Here are 5 related questions:
1. How is the Cauchy-Schwarz inequality derived?
2. What are some applications of Cauchy-Schwarz inequality in physics?
3. Can the equality in Cauchy-Schwarz inequality ever hold?
4. How do inner products differ in real and complex spaces?
5. How can the Cauchy-Schwarz inequality be extended to matrices?

**Tip:** In complex spaces, always remember to use the complex conjugate in the inner product calculation!

This detailed solution verifies the Cauchy-Schwarz inequality for complex vectors x = (2, 1 + i, i) and y = (2 - i, 2, 1 + 2i). Learn how to compute inner products, magnitudes, and apply this fundamental inequality in vector spaces. Suitable for undergraduate students.

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