Let $A\in\text{Mat}_{n\times n}$ be an invertible matrix such that$\left\Vert A\right\Vert \geq c$ where $\left\Vert \cdot\right\Vert $ is the matrix norm . Is it true that for every $\lambda$ eigenvalue of $A$ we have $\left|\lambda\right|\geq c$?
I know that $\left\Vert A\right\Vert \geq\left|\lambda\right|$ but why would $\left|\lambda\right|\geq c$?
Also Im familiar with the norm inequality $\left|Ax\right|\leq\left\Vert A\right\Vert \left|x\right|$ but couldn't use it in favor. Any suggestions?
No. Consider any diagonal matrix of the form $A = \begin{pmatrix} c & 0 \\ 0 & \varepsilon I_{n-1} \end{pmatrix}$ where $I_{n-1}$ is the $n-1 \times n-1$ identity. Then $||A|| = c$, so clearly $||A|| \geq c$, but all its other eigenvalues are $\varepsilon < c$.