Let $A\in\mathbb{R}^N$ be a positive semidefinite matrix with only one zero eigenvaule, i.e. its eigenvalues are given by the following increasing order $0=\lambda_1<\lambda_2\leq...\leq\lambda_N$. Does this imply the following inequality:
$$\lambda_2x^Tx\leq x^TAx\leq\lambda_Nx^Tx$$
Note that $\lambda_2$ is the second eigenvalue not the minimum one. In graph theory, this is true when $A$ is the Laplacian matrix of a graph. I wonder whether this inequality holds for all semidefinite matrices or not.
Take $x=v_1$ the eigenvector corresponding to $\lambda_1 = 0$.
$$ \lambda_2 x^T x \leq x^T A x \leq \lambda_N x^T x\\ \lambda_2 x^T x \leq 0 \leq \lambda_N x^T x\\ $$
But that's not true. $\lambda_2 x^T x > 0$.