Suppose $A$ is a matrix and $x$ a vector of suitable size. I am familiar with the inequality $||Ax||_2\leq ||A||_2 \cdot ||x||_2$. I was wondering if it is possible to find a statement for the other way around for special cases: $||x|| \leq z||Ax||$ with $z$ a suitable scalar. A trivial example would be $z=1$ and $A=I$ or $A$ an orthogonal matrix.
Does anyone know an example?
Use eigen decomposition of the matrix. That is,
$\lambda x=Ax,$
where $x$ is eigenvector and $\lambda$ is corresponding eigenvalue.
Hence any matrix $A$, which has positive eigenvalue will admit the inequity that you have mentioned, i.e.,
$\lambda\Vert x \Vert=\Vert Ax \Vert$ $\to$ $\Vert x \Vert=\frac{1}{\lambda}\Vert Ax \Vert$, or $\Vert x\Vert\leq z\Vert Ax \Vert$, where $z\geq 1/\lambda$
where $\lambda$ and $x$ are eigencomponent of matrix $A$.
I hope that helps