Weighted Inner Product vs Norm Relation

Question

Let $A$ be a matrix with $\lambda_1>0$. Then we know that, \begin{equation} \frac{||x^TAy||}{|||x|||y||} \leq \lambda_1. \end{equation} But is there any lower bound known for the quantity $\frac{||x^TAy||}{|||x|||y||}$?

Answer 1

According to SVD we have $$A=UDV$$where $U$ and $V$ are unitary and $D$ diagonal. Substituting this in the expression we have$$\dfrac{|x^HAy|}{|x|\cdot|y|}=\dfrac{|x^HUDVy|}{|x|\cdot|y|}$$let $v=U^Hx$ and $w=Vy$ therefore $$|v|=|x|\quad,\quad|w|=|y|$$ and we have $$\dfrac{|x^HAy|}{|x|\cdot|y|}=\dfrac{|v^HDw|}{|v|\cdot|w|}$$Without lose of generality, let's assume that $|v|=|w|=1$which leads to $$v_1^2+\cdots+v_n^2=1\\w_1^2+\cdots+w_n^2=1$$therefore we seek to minimize $|v^HDw|$ but the minimum is zero by taking $v=\left(\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}},0,\cdots,0\right)$ and $u=\left(\dfrac{d_2}{\sqrt{d_1^2+d_2^2}},\dfrac{-d_1}{\sqrt{d_1^2+d_2^2}},0,\cdots,0\right)$