Let $A$ be a real matrix with all eigenvalues in the interval $(0,1)$. Show that $$\kappa(A)\le\dfrac{2-\lambda_{\min}(A)}{\lambda_{\min}(A)}$$ where $\kappa(A)$ is $A$'s condition number (maximum singular value $\sigma_{\max}(A)$ divided by minimum singular value $\sigma_{\min}(A)$) and $\lambda_{\min}(A)$ is $A$'s smallest eigenvalue. Inspired by this if we also assume that $N$ is positive definite.
Let $\lambda_{\max}(A)$ be $A$'s biggest eigenvalue. Then $$\dfrac{\lambda_{\max}(A)}{\lambda_{\min}(A)}<\dfrac{1}{\lambda_{\min(A)}}.$$ $\sigma_{\max}(A)\ge\lambda_{\max}(A)$ and $\sigma_{\min}(A)\le\lambda_{\min}(A)$, so $$\dfrac{\lambda_{\max}(A)}{\lambda_{\min}(A)}\le\kappa(A).$$ If $A$ is normal, equality holds and we're done. I don't know how to proceed if $A$ isn't normal.
Let $A=\begin{bmatrix} {1 \over 2} & 10,000 \\ 0 & {1 \over 2} \end{bmatrix}$.
The above formula suggests that an upper bound for the condition number is $3$, but $\|A^{-1} e_1 \| = 2 $ and $\| Ae_2 \| > 10,000$ so we have $\kappa(A) > 20,000$.