Part (a): Determine for which $k \in \mathbb{R}$, the relation defines an inner product

The given mapping is:

$\langle x, y \rangle = x_1y_1 - 2x_1y_2 - 2x_2y_1 + kx_2y_2.$

For this mapping to define an inner product, it must satisfy the following conditions:

1. Bilinearity: The mapping must be linear in both arguments.
2. Symmetry: $\langle x, y \rangle = \langle y, x \rangle$.
3. Positive definiteness: $\langle x, x \rangle > 0$ for all $x \neq 0$.

Step 1: Check symmetry

The expression for $\langle x, y \rangle$ is:

$\langle x, y \rangle = x_1y_1 - 2x_1y_2 - 2x_2y_1 + kx_2y_2.$

Interchanging $x$ and $y$:

$\langle y, x \rangle = y_1x_1 - 2y_1x_2 - 2y_2x_1 + ky_2x_2.$

Rearranging the terms of $\langle y, x \rangle$, we find:

$\langle y, x \rangle = x_1y_1 - 2x_1y_2 - 2x_2y_1 + kx_2y_2.$

This shows that $\langle x, y \rangle = \langle y, x \rangle$, so the mapping is symmetric.

Step 2: Check positive definiteness

To check positive definiteness, compute $\langle x, x \rangle$ for $x = (x_1, x_2)^T$:

$\langle x, x \rangle = x_1^2 - 2x_1x_2 - 2x_2x_1 + kx_2^2.$

Simplify:

$\langle x, x \rangle = x_1^2 - 4x_1x_2 + kx_2^2.$

For $\langle x, x \rangle > 0$ for all $x \neq 0$, the quadratic form must be positive definite. The matrix associated with this quadratic form is:

1 & -2 \\ -2 & k \end{bmatrix}.$$ The quadratic form is positive definite if and only if: 1. The leading principal minor $$a_{11} > 0$$. 2. The determinant of $$A$$ is positive. #### Check the conditions: 1. $$a_{11} = 1 > 0$$, so the first condition is satisfied. 2. The determinant of $$A$$ is: $$\det(A) = (1)(k) - (-2)(-2) = k - 4.$$ For $$\det(A) > 0$$, we require $$k > 4$$. Thus, the mapping defines an inner product if and only if: $$k > 4.$$ --- ### Part (b): Lengths of vectors $$e_1 = (1, 0)^T$$ and $$e_2 = (0, 1)^T$$ Using the inner product with $$k > 4$$, compute the lengths of $$e_1$$ and $$e_2$$: #### Length of $$e_1$$: $$\|e_1\| = \sqrt{\langle e_1, e_1 \rangle}.$$ $$\langle e_1, e_1 \rangle = (1)(1) - 2(1)(0) - 2(0)(1) + k(0)(0) = 1.$$ $$\|e_1\| = \sqrt{1} = 1.$$ #### Length of $$e_2$$: $$\|e_2\| = \sqrt{\langle e_2, e_2 \rangle}.$$ $$\langle e_2, e_2 \rangle = (0)(0) - 2(0)(1) - 2(1)(0) + k(1)(1) = k.$$ $$\|e_2\| = \sqrt{k}.$$ #### Check if $$e_1$$ and $$e_2$$ are orthogonal: Compute $$\langle e_1, e_2 \rangle$$: $$\langle e_1, e_2 \rangle = (1)(0) - 2(1)(1) - 2(0)(0) + k(0)(1) = -2.$$ Since $$\langle e_1, e_2 \rangle = -2 \neq 0$$, the vectors are **not orthogonal**. --- ### Final Results: 1. The mapping defines an inner product if $$k > 4$$. 2. For $$k > 4$$: - $$\|e_1\| = 1$$, - $$\|e_2\| = \sqrt{k}$$, - $$e_1$$ and $$e_2$$ are not orthogonal. --- Do you want further clarifications or more examples on this? Here are 5 related questions to explore further: 1. What is the geometric interpretation of positive definiteness in an inner product? 2. How does the value of $$k$$ affect the angles between vectors in $$\mathbb{R}^2$$? 3. Can we construct an orthonormal basis for $$\mathbb{R}^2$$ using this inner product? 4. How does this inner product generalize to higher dimensions? 5. What is the significance of the determinant of the associated matrix in inner product spaces? **Tip:** To verify positive definiteness in practice, always test eigenvalues of the matrix representation—they must all be positive!