To verify Schwarz's inequality (or Cauchy-Schwarz inequality) for two complex vectors $x$ and $y$, we need to check the inequality:

$| \langle x, y \rangle |^2 \leq \langle x, x \rangle \langle y, y \rangle$

where $\langle x, y \rangle$ is the inner product of the vectors $x$ and $y$.

Step 1: Express the vectors $x$ and $y$

Given:

• $x = (1 + i, 2i)$
• $y = (3i \cdot 0.2 + 3i, 4) = (0.6i + 3i, 4) = (3.6i, 4)$

So we have: $x = (1 + i, 2i), \quad y = (3.6i, 4)$

Step 2: Compute the inner product $\langle x, y \rangle$

The inner product of two complex vectors $x = (x_1, x_2)$ and $y = (y_1, y_2)$ is defined as:

$\langle x, y \rangle = x_1 \overline{y_1} + x_2 \overline{y_2}$

where $\overline{y_1}$ and $\overline{y_2}$ are the complex conjugates of $y_1$ and $y_2$.

For $x = (1 + i, 2i)$ and $y = (3.6i, 4)$:

• $\overline{y_1} = -3.6i$ (since $3.6i$ becomes $-3.6i$ when conjugated)
• $\overline{y_2} = 4$

Now calculate:

$\langle x, y \rangle = (1 + i)(-3.6i) + (2i)(4)$

First term: $(1 + i)(-3.6i) = 1(-3.6i) + i(-3.6i) = -3.6i + 3.6$ Second term: $(2i)(4) = 8i$

So: $\langle x, y \rangle = (-3.6i + 3.6) + 8i = 3.6 + 4.4i$

Step 3: Compute the magnitudes $\langle x, x \rangle$ and $\langle y, y \rangle$

For $\langle x, x \rangle$:

$\langle x, x \rangle = (1 + i)\overline{(1 + i)} + (2i)\overline{(2i)}$ $\overline{(1 + i)} = 1 - i, \quad \overline{(2i)} = -2i$ $\langle x, x \rangle = (1 + i)(1 - i) + (2i)(-2i)$

First term: $(1 + i)(1 - i) = 1^2 - i^2 = 1 + 1 = 2$ Second term: $(2i)(-2i) = -4i^2 = 4$

So: $\langle x, x \rangle = 2 + 4 = 6$

For $\langle y, y \rangle$:

$\langle y, y \rangle = (3.6i)(-3.6i) + (4)(4)$

First term: $(3.6i)(-3.6i) = -12.96i^2 = 12.96$ Second term: $(4)(4) = 16$

So: $\langle y, y \rangle = 12.96 + 16 = 28.96$

Step 4: Verify Schwarz's inequality

We need to check if:

$| \langle x, y \rangle |^2 \leq \langle x, x \rangle \langle y, y \rangle$

First, calculate $| \langle x, y \rangle |^2$:

$| \langle x, y \rangle | = | 3.6 + 4.4i | = \sqrt{(3.6)^2 + (4.4)^2} = \sqrt{12.96 + 19.36} = \sqrt{32.32}$ $| \langle x, y \rangle |^2 = ( \sqrt{32.32} )^2 = 32.32$

Now, calculate $\langle x, x \rangle \langle y, y \rangle$:

$\langle x, x \rangle \langle y, y \rangle = 6 \times 28.96 = 173.76$

Since $32.32 \leq 173.76$, Schwarz's inequality holds.

Conclusion

Schwarz's inequality is verified for the given vectors $x = (1 + i, 2i)$ and $y = (3.6i, 4)$.

Would you like more details or have any questions about this process?

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Here are 5 related questions:

1. How is the inner product of two vectors defined for real and complex numbers?
2. What geometric meaning does the Cauchy-Schwarz inequality have?
3. How do complex conjugates affect vector inner products?
4. What is the significance of verifying inequalities in linear algebra?
5. How does Schwarz's inequality apply to functional spaces in advanced mathematics?

Tip: When dealing with complex numbers, always remember to take the complex conjugate of the second vector in the inner product definition.