What is the solution for these two equalities $ x = \lambda(Ax) $ and $ x^TAx=1 $?

Question

Let $A \in \mathbb{R}^{4 \times 4}$ be a symmetric matrix with two distinct positive eigenvalues and other eigenvalues of $A$ are nonpositive.

What is the solution for $x$ when $\lambda >0$

$$ x = \lambda(Ax) $$ and $$ x^TAx=1 $$

Answer 1

Suppose that $x=\lambda Ax$, then $$\lambda Ax-x=0$$ $$\lambda(Ax-\lambda^{-1}x)=0$$ $$(A-\lambda^{-1}I)x=0$$ So $x$ is an eigenvector with eigenvalue $\lambda^{-1}>0$. Moreover, $$x^T Ax=\lambda^{-1}x^T x=1 \Rightarrow x^T x=||x||^2=\lambda$$

So any solutions, $x$, must be eigenvectors with (positive) eigenvalue $\lambda^{-1}$ and norm $||x||=\sqrt{\lambda}$. If the unit eigenvectors of $A$ are $v_1$ and $v_2$ with the corresponding positve eigenvalues $\mu_1$ and $\mu_2$, then the solutions to your system of equations are: $$(x,\lambda) \in \{(\sqrt{\mu_1^{-1}}v_1,\mu_1^{-1}),(\sqrt{\mu_2^{-1}}v_1,\mu_2^{-1}) \}$$