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A relation between the 2-norm condition number of a matrix $A$ and its eigenvalues is stated as $$ \kappa_2(A) \geq {\max{|\lambda_i(A)|} \over \min {|\lambda_i(A)|} } \tag{1} $$ where $A$ is not necessarily symmetric.

If we take for instance $$ A = \left[ \matrix{ 1 & 2 \cr 3 & 4 \cr} \right], $$ then $$ \kappa_2(A) = 14.9330, \ \max |\lambda_i(A)| = 5.3723, \min |\lambda_i(A)| = 0.3723 $$ and (1) can be verified as $$ {\max |\lambda_i(A)| \over \min |\lambda_i(A)| } = 14.4300 $$

If $A$ is symmetric, then we can use its special properties and in fact, the result (1) holds as an equality.

$$ \kappa_2(A) = {\sigma_\max(A) \over \sigma_\min(A)} = {|\lambda_\max(A)| \over |\lambda_\min(A)|} $$ (for symmetric matrices)

For the general case, how to establish the result (1)?

I attempted a proof using the property that $$ \rho(A) \leq \Vert A \Vert_2 $$ where $\rho(A)$ is the spectral radius of $A$.

This shows that $$ \max|\lambda_i(A)| \leq \Vert A \Vert_2 \tag{2} $$ and next we need to show that $$ {1 \over \min|\lambda_i(A)|} \leq \Vert A^{-1} \Vert_2 $$

We also note that $$ \rho\left( A^{-1} \right) \leq \Vert A^{-1} \Vert_2 $$

As noted in the comments below, the eigenvalues of $A^{-1}$ are the reciprocals of the eigenvalues of $A$.

Hence, it is immediate that $$ {1 \over \min{ | \lambda_i(A) |}} = \rho\left( A^{-1} \right) \leq \Vert A^{-1} \Vert_2 \tag{3} $$

Combining (2) and (3), the result (1) follows, viz.

$$ {\max |\lambda_i(A)| \over \min |\lambda_i(A)|} \leq \kappa_2(A) $$

In fact, this result holds true for the condition number of $A$ ($\kappa(A)$) with respect to any operator norm of $A$, since $\rho(A) \leq \Vert A \Vert$ for any operator norm.

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The eigenvalues of $A^{-1}$ are the reciprocals of the eigenvalues of $A$. So $\frac{1}{\min |\lambda_i(A)|}=\max |\lambda_i(A^{-1})|$; thus your property finishes the proof.

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  • $\begingroup$ Thanks for the great help!! $\endgroup$
    – Sundar
    Commented Apr 14, 2022 at 11:54
  • $\begingroup$ The result $\rho(A) \leq \Vert A \Vert$ holds for any operator norm of $A$. The same proof can be applied to establish a lower bound for the condition number of a matrix with respect to any operator norm in terms of the eigenvalues of $A$ (symmetric or non-symmetric). Thanks! $\endgroup$
    – Sundar
    Commented Apr 14, 2022 at 12:53
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    $\begingroup$ @Dr.Sundar Indeed the statement works for any matrix norm, as long as it's induced by some vector norm. $\endgroup$
    – Ian
    Commented Apr 14, 2022 at 13:31

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