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!