Find solution of optimisation problem $\min_{x} \; x^TAx$ with constraint $\Vert x \Vert^2 \leq 1$

Question

I have to find the solution of the following problem : $A \in R^{n\times n}$ is a symetric positive define matrix $$ \begin{cases} \min_{x} & x^TAx \newline s.c & \Vert x \Vert^2 \leq 1 \end{cases} $$ With the lagrangian : $L(x, \sigma) = x^TAx + \sigma(\Vert x \Vert^2-1) $
I found the KKT conditions : $$ \begin{equation} \begin{cases} 2Ax + 2\sigma x = 0 \newline x^Tx - 1 \leq 0 \newline \sigma (x^Tx-1) = 0 \newline \sigma \geq 0 \end{cases} \end{equation} $$

But I don't know how to find $x^*$ that satisfies the problem.

Answer 1

It is known that

$ \lambda_{Min}(A) \| x \|^2 \le x^T A x \le \lambda_{Max}(A) \| x \|^2 $

Equality in the left inequality occurs when $x$ is an eigenvector corresponding to the minimum eigenvalue of $A$, and equality in the right inequality occurs when $x$ is an eigenvector corresponding to the maximum eigenvalue of $A$.

In addition to the above, we have $ \| x \|^2 \le 1 $

Hence,

$ x^T A x \ge \lambda_{Min}(A) \| x \|^2 $

Now we have two cases:

1. If $\lambda_{Min}(A) \ge 0 $ then the minimum of $ x^T A x $ is $0$ at $x = 0$.

2. If $\lambda_{Min}(A) \lt 0 $ then the minimum of $x^T A x $ is $ \lambda_{Min}(A)$ , and this minimum occurs with $x$ being the unit eigenvector corresponding to the minimum eigenvalue of $A$.