Finding the minimum of Condition number for this matrix

Question

Suppose $A=\left[ {\begin{array}{*{20}{c}} {0.1\alpha }&{0.1\alpha }\\ 1&{1.5} \end{array}} \right]$. How can we find minimum of condition number $k(A)=\Vert A\Vert \Vert A^{-1} \Vert$ (Assume $\infty$-Norm)? And for what value of $\alpha$?

Answer 1

The matrix $A$ is singular if and only if $\alpha = 0$. Suppose $\alpha \neq 0$ and assume the infinity norm. Thus,

$$\|A\|_{\infty} =\begin{cases} -0.2\alpha, & \mbox{if } \alpha \leq -12.5 \\ 2.5, & \mbox{if } -12.5 < \alpha < 12.5\\ 0.2\alpha, & \mbox{if } \alpha \geq 12.5 \end{cases}.$$

We have \begin{equation} A^{-1} = \left [ \begin{array}{cc} 30/\alpha & -2 \\ -20/\alpha& 2\\ \end{array} \right] \end{equation} and $$\|A^{-1}\|_{\infty} =\begin{cases} -\frac{30}{\alpha} + 2, & \mbox{if } \alpha < 0 \\ \frac{30}{\alpha} +2, & \mbox{if } \alpha > 0 \end{cases}.$$ Hence,

$$k(A) =\begin{cases} -0.4\alpha +6, & \mbox{if } \alpha \leq -12.5 \\ 5-\frac{75}{\alpha}, & \mbox{if } -12.5 < \alpha < 0\\ 5+\frac{75}{\alpha}, & \mbox{if } 0 < \alpha < 12.5\\0.4\alpha + 6, & \mbox{if } \alpha \geq 12.5 \end{cases}.$$

The minimal value of $k(A)$ is $11$ for $\alpha = \pm 12.5.$