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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.

Show that the following function defines an inner product on V2(R) where x = (x1, x2) and y = (y1, y2) and ⟨x, y⟩ = 6x1y1 + 7x2y2.

This problem explores how to prove that a given function defines an inner product on V2(R) with vectors x = (x1, x2) and y = (y1, y2). Learn about the properties of inner products, including bilinearity, symmetry, positivity, and definiteness. Ideal for undergraduate students studying linear algebra.

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