Show that the optimum value for the problem is $\frac{\lambda_1}{2}$ Let $G\in\mathbb{R}^{n\times n}$ be a symmetric matrix with eigenvalues $\lambda_1\leq\lambda_2\leq\cdots\leq\lambda_n$.
Let the problem be:
$$\text{minimize}\quad\frac{1}{2}\vec{x}^TG\vec{x}$$
$$\text{s.t.} \quad \vec{x}^T\vec{x}=1$$
Show that the optimum value for the problem is $\frac{\lambda_1}{2}$
My try:
The Lagrangian is: $$L=\frac{1}{2} \vec{x}^T G\vec{x} + \mu(\vec{x}^T \vec{x}-1)$$
Then if we take the gradient with respect to $\vec{x}$ and set it equal to the zero vector:
$$\nabla_x L= G\vec{x}+2\mu\vec{x}=\vec{0}$$
Then $$\vec{x}=G^{-1}(-2\mu\vec{x})$$
I'm not sure on how to solve for $\vec{x}$ here. Any suggestions would be great!

Hint: $Gx=-2\mu x$ implies that $x$ is an eigenvector of $G$.
