onsider the following problem:
$$\min_{x \in \mathbb{R}^4} x^Tx$$
over $C=\{x \in \mathbb{R}^4 \mid x^TAx \geq 1\}$ where $A \in \mathbb{R}^{4 \times 4}$ is a symmetric matrix with two distinct positive eigenvalues and other eigenvalues of $A$ are nonpositive.
Find KKT point of the problem.
My try:
We can show that every Fritz-John point is a KKT point so we can write the following system of equlity and inequality to find the minimizer. Let $L(x, \lambda)=x^Tx+\lambda (1-x^TAx)$ be the Lagrangian. Then,
$$ 2x + \lambda(-2Ax)=0 \tag{1} $$ And,
$$ \lambda \geq 0 , x^TAx \geq 1, \lambda (1-x^TAx)=0 \tag{2} $$
From (1) we know that $\lambda$ cannot be zero because otherwise $x=0$ which violates the constraint. So it is $\lambda > 0$ and we have
$$ x = \lambda(Ax) $$ And,
$$ x^TAx=1 $$
How can I proceed?
From $x=\lambda Ax$ you know that $x$ is an eigenvector of $A$ (since $Ax = (1/\lambda) x$). Let $x$ be an eigenvector with eigenvalue $\mu$. Since $x^T A x = 1$, $\mu x^Tx = 1$, so $x^Tx = 1/\mu$. The objective value is therefore minimized by the eigenvector with the largest eigenvalue.