2-norm of matrix

Question

How to prove that for a symmetric matrix A with eigenvalues $\lambda_1 \leq \lambda_ 2, ... \leq \lambda_n$ it holds that $$\vert\vert A \vert\vert = \text{max}(-\lambda_1, \lambda_n)$$ where $\vert\vert \cdot\vert\vert$ denotes the 2-norm? I am familiar with the usual matrix norm, i.e. $\text{max}(\lambda_i)$, but the $-\lambda_1$ confuses me.

Answer 1

If $v_i$ are the corresponding orthonormal eigenvectors of $A$ then you have

$$ ||Av||^2 = \left< A\left( \sum_{i=1}^n \left< v, v_i \right>v_i \right), A \left( \sum_{i=1}^n \left< v, v_j \right> v_j\right) \right> = \left< \sum_{i=1}^n \left<v, v_i \right> \lambda_i v_i, \sum_{j=1}^n \left< v, v_j \right> \lambda_j v_j \right> \\ = \sum_{i=1}^n \lambda_i^2 \left<v, v_i \right>^2 \leq \max_{i} \lambda_i^2 \sum_{i=1}^n \left<v, v_i \right>^2 = \max(-\lambda_1, \lambda_n)^2 ||v||^2. $$

In the equations above we used the fact that $v_i$ is an orthonormal basis and so $\left< v_i, v_j \right> = \delta_{ij}$ and the fact that the eigenvalues are ordered and so $\max_{i} {\lambda_i^2} = \max (\lambda_1^2, \lambda_n^2) = \max(-\lambda_1, \lambda_n)^2. $