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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.

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    $\begingroup$ $max(|\lambda_i|)=max(-\lambda_1,\lambda_n)$ $\endgroup$ – Paul Jun 1 '16 at 18:59
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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. $

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  • $\begingroup$ I understand all of your proof, but can you explicitly argue how you got from the definition of the matrix 2 norm ($||A||_2:=\max_{||x||=1}{||Ax||_2}$) to $||A||_2^2=||Av||^2$ where $v$ is a orthonormal eigenvector of $A$? $\endgroup$ – NoseKnowsAll Jun 1 '16 at 20:22
  • $\begingroup$ @NoseKnowsAll: Not as to speak for levap, but the square is just to kill the root and simplify calculations I guess. Further, there is an orthonormal eigenvector of A because it is symmetric and thus orthogonal diagonalizable. Thanks for the proof, levap! $\endgroup$ – Taufi Jun 1 '16 at 20:46
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    $\begingroup$ @NoseKnowsAll: The calculation above works for any $v \in \mathbb{R}^n$, not just for an eigenvector! If you take $v = x$ with $||x|| = 1$ then you'll see that $||A|| \leq \max(-\lambda_1, \lambda_n)$. However, this maximum is actually attained (either for $v = v_1$ or $v = v_n$). $\endgroup$ – levap Jun 1 '16 at 21:00

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