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Let's play with algebra

What other ways can I express this inequality: $\|\mathbf {Ad} \|_{2}\geq \epsilon \|\mathbf {d} \|_{2}$ where $\mathbf {A}$ is a matrix, $\mathbf {d}$ is a vector, and $\epsilon$ is a positive scalar? Is there something I can say about the relationship of $\mathbf {d}$ to the nullspace of $\mathbf {A}$? I also tried playing with the triangle inequality but that didn't get me anywhere. Is there a way to express the equation linearly (strangely enough, in the elements of $\mathbf {A}$) or otherwise well to use as a constraint in an optimization problem in $\mathbf {A}$?

What if $\mathbf {d}$ is the gradient of a function? I have this idea that if we're projecting a function $f(\mathbf {y} )$ into a new function $g(x)=f(\mathbf {Ax} )$, ensuring that $\|\mathbf {A} \nabla f(y)\|_{2}\geq \epsilon \|\nabla f(y)\|_{2}$ will ensure that critical points of $g(\mathbf {x} )$ will also be critical points of $f(\mathbf {y} )$, and that generally if we're performing optimization we can make some progress in minimizing $f(\mathbf {y} )$ by performing optimization over $g(\mathbf {x} )$ and then setting $\mathbf {y} _{*}=\mathbf {Ax} _{*}$, then maybe starting again. Any graphical or intuitive understanding of what this inequality would mean or a better one to choose to accomplish that purpose would be helpful.

If I have second order information, would I do better setting $\mathbf {d} =\mathbf {H} ^{-1}\nabla f(y)$ (a Newton's method step) where $\mathbf {H}$ is the Hessian matrix or some approximation to it?

I know my question is confusing. Please just answer whatever small parts you can and go off on any tangents you think might be helpful. 18.252.5.40 08:30, 9 August 2006 (UTC)

The obvious first suggestion is to read our article on matrix norms. --KSmrq^T 09:25, 9 August 2006 (UTC)