Question

Let the superscript \( T \) represent the transpose operation. Consider the function \[ f(x) = \frac{1}{2} x^T Q x - r^T x, \] where \( x \) and \( r \) are \( n \times 1 \) vectors and \( Q \) is a symmetric \( n \times n \) matrix. The stationary point of \( f(x) \) is

• A. \( Q^T r \)
• B. \( Q^{-1} r \)
• C. \( r \, r^T \)
• D. \( r \)

Hint:

To find the stationary point of a quadratic function, take the derivative of the function, set it equal to zero, and solve for \( x \).

Answer 1

The Correct Option is B

We are given the function \( f(x) = \frac{1}{2} x^T Q x - r^T x \), where \( Q \) is a symmetric matrix and \( x \) and \( r \) are \( n \times 1 \) vectors. To find the stationary point of \( f(x) \), we take the gradient of \( f(x) \) and set it equal to zero: \[ \nabla f(x) = \frac{\partial}{\partial x} \left( \frac{1}{2} x^T Q x - r^T x \right). \] Step 1: Compute the derivative Using matrix calculus:
- The derivative of \( \frac{1}{2} x^T Q x \) with respect to \( x \) is \( Qx \),
- The derivative of \( -r^T x \) with respect to \( x \) is \( -r \).
Thus, the gradient is: \[ \nabla f(x) = Qx - r. \] Step 2: Set the gradient equal to zero To find the stationary point, we set \( \nabla f(x) = 0 \): \[ Qx - r = 0 $\Rightarrow$ Qx = r. \] Step 3: Solve for \( x \) Since \( Q \) is invertible (as it is symmetric and typically assumed to be positive definite), we can solve for \( x \): \[ x = Q^{-1} r. \] Thus, the stationary point is \( x = Q^{-1} r \), corresponding to Option (B).
Final Answer: (B) \( Q^{-1} r \)