I would like to show that the minimum of $J(x)=\langle Ax,x\rangle$ (where $A$ is a symmetric $n$ by $n$ matrix) on $K=\{x\in\mathbb{R}^{n} : F(x) = 1 -\lVert x \rVert^{2} = 0 \}$ is the eigenvector associated to the smallest eigenvalue of $A$.
To do this I use the Lagrange theorem, we now that $K$ is compact and $J$ is continuous so there exists a solution to this problem, let’s say $u$. Now from the Lagrange necessary condition we know that there exists $\lambda$ such that
$$ J’(u)(h) + \lambda F’(u)(h) = 0\quad\forall h\in\mathbb{R}^{n} $$
Here we have $J’(u)(h) = 2\langle Au, h\rangle$ and $F’(u)(h) = 2\langle x,h\rangle$.
Thus we have
$$ \forall h\in\mathbb{R}^{n}, \langle Au -\lambda u, h\rangle = 0 \implies Au = \lambda u $$
From this we have indeed that $u$ is the eigenvector associated to $\lambda$. However I do not see how to prove that $\lambda$ is the smallest egeinvalue of $A$ ?
If you have any hints (instead of full answer, in order to work a little bit by myself) do not hesitate to share it please.
Thank you
Just use the fact that you minimize over the unit sphere and the connection between dot product and norm. $$\langle Au,u\rangle=\langle \lambda u,u\rangle=\lambda \langle u,u\rangle=\lambda \|u\|^2 =\lambda $$