Let $A$ be an invertible matrix of size $n \times n$. Let $v$ be an arbitrary unit vector in $\mathbb{R}^n$, i.e., $\|v\|_2=1$. Then there exists a unit vector $u \in \mathbb{R}^n$ such that $\|A^{-1}u\|_2 = \|A^{-1}\|_2$.
I am trying to show the following inequality $$ \|A^{-1}v\|_2 \ge \|A^{-1}\|_2|u^Tv|. $$
Here is my try.
We can write $$ v = (u^Tv)u + (v-(u^Tv)u). $$ Thus $$ A^{-1}v = (u^Tv)A^{-1}u + A^{-1}(v-(u^Tv)u).$$ If somehow the first and the second terms in the RHS are orthogonal, I can write $$ (*) \quad \|A^{-1}v\|_2^2 = \|(u^Tv)A^{-1}u \|_2^2 + \|A^{-1}(v-(u^Tv)u\|_2^2 \ge \|(u^Tv)A^{-1}u \|_2^2 $$ which gives the desired inequality $$ \|A^{-1}v\|_2 \ge \|(u^Tv)A^{-1}u \|_2 = |u^Tv|\|A^{-1}u\|_2 = |u^Tv|\|A^{-1}\|_2.$$ However, (*) is not true in general. So I am stucked on here.
Any comments or suggestions will be very appreciated. Thanks in advance.
By a change of (orthonormal) basis, we may assume that $A^{-1}=Q\Sigma$ is a singular value decomposition, where $Q$ is real orthogonal, $\Sigma=\operatorname{diag}(\sigma_1,\ldots,\sigma_n)$ is the singular value matrix and $u=e_1=(1,0,\ldots,0)^T$. Then the inequality in question reduces to $\|\Sigma v\|_2\ge\sigma_1|e_1^Tv|$, which is trivial.