Let a is fixed $f:R^{n^2}\times R^n\to R$ such that $f(A,u)=u^TAu+a^Ta$ where $A$ is a symmetric, invertible matrix. Find $u$ such that $f$ attains the minimum for that $A$.
I know that suppose f attain minima then at that point partial derivative is o
$f_u=2u^TA$ its partial derivative
as A is invertible implies $u=\bar 0$
So mimima occur at 0.
But is it possible to find unit vector that minimises f?
That is I wanted to find unit vector u such that for any unit vector v $f(u)\leq f(v)$?
Any Help will be appreciated
$\left\{\begin{array} . f \to \min\limits_{u} \\ s.t. \ u^T u = 1 \end{array} \right. $
Let's use the Lagrangian: $L = u^T A u + a^Ta - \lambda \cdot (u^Tu - 1)$. Necessary condition of minimum - $\nabla_{u} L = 0$.
$\nabla_{u} L = 2Au - 2\lambda u = 0 \Rightarrow Au = \lambda u \Rightarrow$ u should be the (normilized) eigenvector, corresponding to the minimum eigenvalue of matrix A.