Given smallest eigen value, give a bound on largest element of $A^{-1}$

Question

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 $A^{-1}=A^T$

Answer 1

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.)

Answer 2

If a matrix $M$ is symmetric then the operator norm of $M$ is given by

$$\|M\| = \max_{\lambda \in \sigma(M)} |\lambda|$$

The $(i,j)$-th entry of a matrix $M$ is given by $\langle Me_j, e_i\rangle$ where $\{e_1, \ldots, e_n\}$ is the canonical basis.

Therefore $$\left|\langle Me_i, e_j\rangle\right| \stackrel{CSB}{\le} \|Me_i\|\|e_j\| \le \|M\|$$

If your matrix $A$ is symmetric, then $A^{-1}$ is also symmetric. Furthermore, its spectrum is given by $$\sigma(A^{-1}) = \frac{1}{\sigma(A)} = \left\{\frac1\lambda : \lambda \in \sigma(A)\right\}$$

If $\lambda_{\text{min}}$ is the smallest (by absolute value) eigenvalue of $A$, then $\displaystyle\frac1{\lambda_{\text{min}}}$ is the largest (by absolute value) eigenvalue of $A^{-1}$.

Hence:

$$\left|\langle A^{-1}e_i, e_j\rangle\right| \le \|A^{-1}\| = \frac{1}{|\lambda_\text{min}|}$$

Answer 3

There is no limit. Let $\lambda_{min} \lt 0$ and $A=\begin {pmatrix} \lambda_{min}&0\\0&\epsilon \end {pmatrix}$. We can make $\epsilon$ as small as we want in absolute value and $A^{-1}=\begin {pmatrix} 1/\lambda_{min}&0\\0&1/\epsilon \end {pmatrix}$