Given smallest eigen value, give a bound on largest element of $A^{-1}$ I've been looking at this question for hours now, but I can't grasp it: 
Let A be a symmetric matrix with minimum eigenvalue $\lambda_{\text{min}}$ . Give a bound on the largest element of $A^{−1}$.
I was looking at the spectral decomposition where $A=VDV^T$ with D a diagonal matrix, however I don't think this is the rightway since for this case is not always true that $A^{-1}=A^T$

The $(i,j)$ entry of a matrix $B$ can be written as $e_i^\top B e_j$ where $e_i$ is the $i$th standard basis vector.
By Cauchy-Schwarz, we have $|B_{ij}| = |e_i^\top B e_j| \le \|B e_j\| \le \sigma_{\max}(B)$ where $\sigma_{\max}(B)$ is the largest singular value of $B$. If $B$ is symmetric, this is the largest eigenvalue of $B$ in absolute value.
We apply the above with $B=A^{-1}$.
If $\lambda_{\min}$ is the smallest eigenvalue of $A$ in absolute value, then the maximum singular value of $A^{-1}$ is $|\lambda_{\min}|^{-1}$.
(Note that we must have $|\lambda_{\min}|>0$ in order for $A$ to be invertible.)
