Cauchy-Schwarz inequality on matrix-vector multiplication Is it possible to choose a vector $b$, which, when multiplied by a matrix $A$ satisfies that $\|Ab\|=\|A\|\|b\|$.
The matrix-norm is the usual 2-norm: $\|A\|=\lambda_\max$ such that $\lambda_\max$ is the largest number that satisfies $A^*A-I\lambda$ is singular.
 A: The answer is yes: consider maximising
$$ L = \lVert Ab \rVert^2-\lambda(\lVert b \rVert^2-1). $$
Expanding $L$ out gives
$$ L = b^* A^* A b - \lambda (b^* b -1), $$
and then you can carry out variation with respect to $b,b^*$ and $\lambda$ to show that the solution to this has $\lambda = \lVert A \rVert$.
Since this is a finite-dimensional maximisation problem over a compact set and $L$ is continuous, it certainly has this maximum.

Alternative: consider $ B := A^* A - \lambda_{\text{max}} I $. This is singular, and hence $\ker{B}$ contains a nonzero vector, $b$, say. But then
$$ 0 = B b = (A^* A - \lambda_{\text{max}} I)b, $$
so $b$ is an eigenvector of $A^* A$ with eigenvalue $\lambda_{\text{max}}$,
$$ A^* A b = \lambda_{\text{max}}b. $$
Multiplying on the left by $b^*$, we find
$$ \lVert Ab \rVert^2 = b^* A^* A b = \lambda_{\text{max}} b^* b = \lVert A \rVert^2 \lVert b \rVert^2, $$
as required.
(I think your notation's a bit off: I think you need a $\lambda^2$ in your definition of $\lambda_{\text{max}}$.)
