Question

Consider the problem of maximizing \[ z = \begin{pmatrix} x_1 & x_2 & x_3 \end{pmatrix} \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 1 & 2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \] subject to \[ \begin{pmatrix} x_1 & x_2 & x_3 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 1, \] where \( \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \in \mathbb{R}^3 \).

Then the maximum value of \( z \) is ______. (Answer in integer)

Hint:

The maximum of a quadratic form x^T A x over unit vectors equals the largest eigenvalue of A, attained at the matching unit eigenvector.

Answer 1

Correct Answer: 3

Step 1: Recognise the form of the problem.
Write \( A = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 1 & 2 \end{pmatrix} \) and \( x = (x_1,x_2,x_3)^T \). The problem is to maximize \( z = x^T A x \) subject to \( x^T x = 1 \), that is, over all unit vectors in \( \mathbb{R}^3 \). \( A \) is symmetric.

Step 2: Use Lagrange multipliers.
Set \( L = x^TAx - \lambda(x^Tx - 1) \). Taking the gradient with respect to \( x \) and setting it to zero gives
\[ 2Ax - 2\lambda x = 0 \implies Ax = \lambda x \]
So the critical points of \( z \) on the unit sphere are exactly the unit eigenvectors of \( A \), and at such a point \( z = x^TAx = \lambda x^Tx = \lambda \). This means the maximum value of \( z \) is the largest eigenvalue of \( A \).

Step 3: Find the eigenvalues of \( A \).
Solve \( \det(A - \lambda I) = 0 \).
\[ \det \begin{pmatrix} 2-\lambda & 0 & 0 \\ 0 & 2-\lambda & 1 \\ 0 & 1 & 2-\lambda \end{pmatrix} = (2-\lambda)\left[(2-\lambda)^2 - 1\right] = 0 \]
One root is \( \lambda = 2 \) directly. The other factor gives \( (2-\lambda)^2 = 1 \), so \( 2-\lambda = \pm 1 \), which means \( \lambda = 1 \) or \( \lambda = 3 \).

Step 4: Pick the largest eigenvalue.
The three eigenvalues of \( A \) are \( 1, 2, 3 \). The largest is \( 3 \), so the maximum value of \( z \) on the unit sphere is \( 3 \).

Final Answer:
The maximum value of \( z \) is \( 3 \). \[ \boxed{3} \]