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I saw this inequality and was not sure how to prove it: let $x$ be a unit vector such that $\Ax\_2=\A\_2$, and moreover, $A$ is invertible and $x$ is unique up to sign; that is, the spectral norm of $A$ is achieved by $x$ and $x$ on the unit sphere. Then for any unit vector $y$, we have $$\Ay\_2\geq \ A\_2\vert x^Ty\vert.$$
This would follow from $\ Ay\_2\geq \ A(xx^T)y\_2$, but I don't see why that should be true. It seems to make intuitive sense (by projecting onto the maximizing vector first, the norm might fall by more than is gained than by the fact $A$ blows it up more). The context was that $y$ and $A$ are random, but this fact seems to have been stated without appeal to this setting. Any ideas/am I missing something obvious? Thanks!
Edit: Added the extra conditions on $A$ and $x$. Sorry about that!
Write $y = a_1 v_1 + ... + a_n v_n$ where $v_1,...,v_n$ is an orthonormal basis of eigenvectors of $A^T A$ for eigenvalues $\lambda_1,...,\lambda

Write $y = a_1 v_1 + ... + a_n v_n$ where $v_1,...,v_n$ is an orthonormal basis of eigenvectors of $A^T A$ for eigenvalues $\lambda_1,...,\lambda_n$ and we can take $v_1 = x$; so $\A\_2 = \lambda_1$. Then $$\Ay\^2_2 = \lambda_1^2 a_1^2 + ... + \lambda_n^2 a_n^2 \ge \lambda_1^2 a_1^2 = \A\_2^2 x^T y^2.$$
20171203 03:58:33