I'm trying to optimize $\vec v A \vec v^T$ for its maximum under the constraint $\vec v \vec v^T \ = \ 1$. However, I'm not allowed to turn it into a form of Lagrangian function $L(\lambda,v,A)$. How can I solve it by expanding these equation? Thanks.
$$ \left\{ \begin{array}{l} \nabla (\vec v A \vec v^T) \ = \ \lambda \nabla (\vec v \vec v^T) \\ \vec v \vec v^T \ = \ 0 \end{array}\right. $$ where $v$ is a column vector and $A$ is a symmetric matrix. I want to solve it by expanding, instead of constructing the Lagrange function $L(\lambda,v,A)$.
I guess you are deriving the maximum variance of PCA and have trouble finding partial derivatives. Since $A$ is symmetric matrix, you might apply the conclusion $$ \frac{\partial}{\partial \vec{v}}\left(\vec{v}^{T} A \vec{v}\right)=2 A \vec{v} $$ As for the constraint funtion, $$ \frac{\partial}{\partial \vec v} \left( \vec v^\top \vec v \right) = 2 \vec v $$ Then you can simply put them together $$ 2A\vec v = 2 \lambda \vec v \\ A\vec v = \lambda \vec v $$