The problem asks to show that the given function defines an inner product on $V_2(\mathbb{R})$. We are given the following information:

Vectors:

• $x = (x_1, x_2)$
• $y = (y_1, y_2)$

The function (possibly the candidate inner product) is: $\langle x, y \rangle = 6x_1y_1 + 7x_2y_2$

To prove this function defines an inner product, we need to verify that it satisfies the following four properties of an inner product:

1. Bilinearity:

   • Additivity in the first argument: $\langle x_1 + x_2, y \rangle = \langle x_1, y \rangle + \langle x_2, y \rangle$
   • Homogeneity in the first argument: $\langle \alpha x, y \rangle = \alpha \langle x, y \rangle$

2. Symmetry: $\langle x, y \rangle = \langle y, x \rangle$

3. Positivity: $\langle x, x \rangle \geq 0$

4. Definiteness: $\langle x, x \rangle = 0 \iff x = 0$

Let's check each property for the given function.

1. Bilinearity:

• Additivity: We need to check if $\langle (x_1 + x_1', x_2 + x_2'), y \rangle = \langle x, y \rangle + \langle x', y \rangle$ Substitute into the given formula for the inner product: $\langle (x_1 + x_1', x_2 + x_2'), y \rangle = 6(x_1 + x_1')y_1 + 7(x_2 + x_2')y_2$ Expanding: $= 6x_1y_1 + 6x_1'y_1 + 7x_2y_2 + 7x_2'y_2$ Which simplifies to: $= (6x_1y_1 + 7x_2y_2) + (6x_1'y_1 + 7x_2'y_2) = \langle x, y \rangle + \langle x', y \rangle$ So, additivity holds.

• Homogeneity: We need to check if: $\langle \alpha x, y \rangle = \alpha \langle x, y \rangle$ Substitute $\alpha x = (\alpha x_1, \alpha x_2)$: $\langle \alpha x, y \rangle = 6(\alpha x_1)y_1 + 7(\alpha x_2)y_2 = \alpha(6x_1y_1 + 7x_2y_2) = \alpha \langle x, y \rangle$ So, homogeneity holds.

2. Symmetry:

We need to check if: $\langle x, y \rangle = \langle y, x \rangle$ Using the given function: $\langle x, y \rangle = 6x_1y_1 + 7x_2y_2$ Now compute $\langle y, x \rangle$: $\langle y, x \rangle = 6y_1x_1 + 7y_2x_2 = 6x_1y_1 + 7x_2y_2$ Thus, $\langle x, y \rangle = \langle y, x \rangle$, so symmetry holds.

3. Positivity:

We need to check if: $\langle x, x \rangle \geq 0$ Substitute $x$ into the given function: $\langle x, x \rangle = 6x_1^2 + 7x_2^2$ Since squares of real numbers are non-negative, $6x_1^2 + 7x_2^2 \geq 0$, so positivity holds.

4. Definiteness:

We need to check if: $\langle x, x \rangle = 0 \iff x = 0$ From the positivity check, $\langle x, x \rangle = 6x_1^2 + 7x_2^2$. For this to be zero, both terms must be zero: $6x_1^2 + 7x_2^2 = 0 \implies x_1 = 0 \text{ and } x_2 = 0$ Thus, $x = 0$ is the only solution, so definiteness holds.

Conclusion:

The function $\langle x, y \rangle = 6x_1y_1 + 7x_2y_2$ satisfies all four properties of an inner product, so it defines an inner product on $V_2(\mathbb{R})$.

Do you need more details or have any questions?

5 Related Questions:

1. How can inner products be used in geometry to define angles between vectors?
2. What is the difference between an inner product and a dot product?
3. How does the concept of an inner product extend to higher-dimensional vector spaces?
4. Can we define inner products on vector spaces over complex numbers?
5. What are some applications of inner product spaces in physics or engineering?

Tip:

Always check that the inner product is positive definite, as it's a key property that ensures the inner product gives meaningful geometric interpretations like angles and lengths.