Answer:
To find the Hessian matrix and construct the quadratic form \( z^T H(x) z \), where \( H(x) \) is the Hessian matrix, we need to follow these steps:
1. Calculate the gradient vector of \( y \) with respect to \( x \).
2. Calculate the Hessian matrix of \( y \) with respect to \( x \).
3. Replace \( x \) with the given vector \( z \) in both the gradient vector and the Hessian matrix.
4. Construct the quadratic form \( z^T H(x) z \) using the modified Hessian matrix and the given vector \( z \).
Let's solve each part:
a. \( y = 2x_1 - x_1x_1 \)
1. Gradient vector: \( \nabla y = [2 - 2x_1] \)
2. Hessian matrix:
\[ H(x) = \begin{bmatrix} -2 \end{bmatrix} \]
3. Substitute \( z \) into the gradient vector and Hessian matrix:
\[ \nabla y_z = [2 - 2z_1] \]
\[ H_z(x) = \begin{bmatrix} -2 \end{bmatrix} \]
4. Construct the quadratic form: \( z^T H(x) z \):
\[ z^T H_z(x) z = \begin{bmatrix} z_1 \end{bmatrix} \begin{bmatrix} -2 \end{bmatrix} \begin{bmatrix} z_1 \end{bmatrix} = -2z_1^2 \]
b. \( y = 2x_1 - x_1 + 2x_2 - 4x_2 \)
1. Gradient vector: \( \nabla y = [2 - 2, 2 - 4] = [0, -2] \)
2. Hessian matrix:
\[ H(x) = \begin{bmatrix} 0 & 0 \\ 0 & -4 \end{bmatrix} \]
3. Substitute \( z \) into the gradient vector and Hessian matrix:
\[ \nabla y_z = [0, -2] \]
\[ H_z(x) = \begin{bmatrix} 0 & 0 \\ 0 & -4 \end{bmatrix} \]
4. Construct the quadratic form: \( z^T H(x) z \):
\[ z^T H_z(x) z = \begin{bmatrix} z_1 & z_2 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 0 & -4 \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} = -4z_2^2 \]
c. \( y = x_1^2 - 6x_1x_2 + 2x_2^2 \)
1. Gradient vector: \( \nabla y = [2x_1 - 6x_2, -6x_1 + 4x_2] \)
2. Hessian matrix:
\[ H(x) = \begin{bmatrix} 2 & -6 \\ -6 & 4 \end{bmatrix} \]
3. Substitute \( z \) into the gradient vector and Hessian matrix:
\[ \nabla y_z = [2z_1 - 6z_2, -6z_1 + 4z_2] \]
\[ H_z(x) = \begin{bmatrix} 2 & -6 \\ -6 & 4 \end{bmatrix} \]
4. Construct the quadratic form: \( z^T H(x) z \):
\[ z^T H_z(x) z = \begin{bmatrix} z_1 & z_2 \end{bmatrix} \begin{bmatrix} 2 & -6 \\ -6 & 4 \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} = 2z_1^2 - 12z_1z_2 + 4z_2^2 \]
So, for:
a. \( z^T H(x) z = -2z_1^2 \)
b. \( z^T H(x) z = -4z_2^2 \)
c. \( z^T H(x) z = 2z_1^2 - 12z_1z_2 + 4z_2^2 \)
