How can I solve an optimization problem $x^T A x$ with constraint $x^T x = 1$? 
Let $A \in \mathbb{R}^{n \times n}$ be a positive definite matrix.
\begin{align}
        &\operatorname*{minimize}_{x \in \mathbb{R}^n} & & x^T A x \\
        &\text{subject to}
        &&x^T x = 1
        \end{align}
What is the minimum which fulfills the constraints?

My thoughts


*

*The constraint means that all possible solutions are on a unit sphere.

*A necessary condition (if there were no constraints) would be


$$
    \frac{\partial x^T A x}{\partial x} = 2 A x \overset{!}{=} 0
$$


*

*The Lagrange function is


\begin{align}
\mathcal{L} (x, \lambda) &= x^T A x + \lambda (x^T x - 1)\\
\nabla_x \mathcal{L} &= \nabla_x (x^*)^T A x^* + \lambda \nabla_x ((x^*)^T x^* - 1) \\
 &= 2 \cdot A \cdot x^* + \lambda \cdot 2 \cdot x^* \overset{!}{=} 0\\
\Leftrightarrow 0 &\overset{!}{=} A \cdot x^* + x^*\\
\Leftrightarrow 0 &\overset{!}{=} (A + \lambda I) \cdot x^*\\
\frac{\partial}{\partial \lambda} \mathcal{L} &= x^T x - 1 = 0 
\end{align}
I have no idea if this is correct / how to continue.
 A: $$\mathrm x^T \mathrm A \mathrm x \geq \lambda_{\min} (\mathrm A) \|\mathrm x\|_2^2 = \lambda_{\min} (\mathrm A) > 0$$
because $\|\mathrm x\|_2 = 1$ and $\mathrm A \succ \mathrm O$. The minimum is attained at the intersection of the eigenspace of $\lambda_{\min} (\mathrm A)$ with the unit Euclidean sphere.
A: Sketch: 


*

*Since $A$ is positive-definite, it is symmetric and hence has an orthonormal eigenbasis.  Let $v_1,\cdots,v_n$ be the eigenvectors with corresponding eigenvalues $\lambda_1,\cdots,\lambda_n$.  

*Therefore we can write $x=\sum a_iv_i$ where $\sum a_i^2=1$.  The sum condition comes from the fact that $\langle\sum a_iv_i,\sum a_iv_i\rangle=\sum a_ia_j\langle a_i,a_j\rangle=\sum a_i^2\langle a_i,a_i\rangle=\sum a_i^2$ by orthonormality.

*One can easily calculate $x^TAx$ by noting that $Ax=\sum a_i\lambda_iv_i$ and then taking the inner product as above.  This results in $\sum a_i^2\lambda_i$ by orthonormality, which is minimized/maximized when all but one $a_i$ is zero.
See also Raleigh Quotient.
A: The above approaches are correct, but just to make the link with your solution, here is another way of writing things:
$$
\begin{cases}
\nabla_x \mathcal{L} = 0\\
\nabla_{\lambda} \mathcal{L} = 0\\
\end{cases}
\quad \Rightarrow\quad
\begin{cases}
Ax = \lambda x\\
x^{T}x=1\\
\end{cases}
$$
In other words, unitary eigenvectors of $A$ satisfy the Lagrangian condition.
Note that the set $x^{T}x=1$ is not convex, but it is compact, so $f(x)=x^{T}Ax$ does reach its minimum on $x^{T}x=1$. So our unitary eigenvectors are definitely good candidates here.
Let $\hat{x}$ be a unitary eigenvector of $A$. It follows that 
$$
f(\hat{x})=\hat{x}^{T}A\hat{x}=\lambda x^{T}x=\lambda
$$
We can conclude that the minimum of $f$ is the smallest eigenvalue of $A$, and is reached by its unitary eigenvector.
