Suppose $A$ is a $202 \times 202$ matrix with $\|A\|_2 = 100$ and $\|A\|_f = 101$. Give the sharpest lower bound on the 2-norm condition number $k(A)$.
I know $k(A) = \|A\|\cdot\|A^{-1}\| = 100 \|A^{-1}\|$
I also know that $\|A\| = \sup_x \frac{\|Ax\|}{\|x\|} = 100$
I'm also not sure, but I think that since the norm of $x$ is a scalar, that I may be able to bring it into the fraction so that $\|A\| = \sup_x \left\|A\frac{x}{\|x\|}\right\| = 100$, if this is true then I can restrict $x$ to be a unit vector. Not sure if this would help.
Can anyone explain how to do this problem?
This question is from Numerical Linear Algebra - Trefethen & Bau
$100=\big\Vert A \big\Vert_2 = \sigma_1$
$101^2 = \big\Vert A \big\Vert_F^2 = \sigma_1^2 + \sigma_2^2 + ....+ \sigma_{201}^2 + \sigma_{202}^2$
$k(A) =\frac{\sigma_1}{\sigma_{n}} =\frac{\sigma_1}{\sigma_{202}}= \frac{100}{\sigma_{202}}$
To get a lower bound on $k(A)$ we want to maximize the denominator. So what's the largest value that $\sigma_{202}$ may take?