The condition number $\kappa(A)$ of a matrix $A$ is defined as $\kappa(A) = \| A \| \cdot \| A^{-1} \|$,
where $\left\|A\right\|=\max_{x\neq 0}\frac{\left\|Ax\right\|}{\left\|x\right\|}=\max_{\left\|x\right\|=1}\left\|Ax\right\|$.
I want to show for a non-zero scalar $c$ that
$\kappa (cA) = \kappa(A)$.
So we can start writing
\begin{equation}
\begin{split}
\kappa (cA) = \| cA \| \cdot \| cA^{-1} \| &= \max_{x\neq 0}\frac{\left\|cAx\right\|}{\left\|x\right\|} \cdot \max_{x\neq 0}\frac{\left\|cA^{-1}x\right\|}{\left\|x\right\|} \\
&= \max_{x\neq 0}\frac{|c| \cdot \left\|Ax\right\|}{\left\|x\right\|} \cdot \max_{x\neq 0}\frac{|c| \cdot \left\|A^{-1}x\right\|}{\left\|x\right\|} \\
&= \max_{x\neq 0}\frac{|c| \cdot \left\|Ax\right\|}{\left\|x\right\|} \cdot \left( \min_{x\neq 0}\frac{\left\|x\right\|}{|c| \cdot \left\|A^{-1}x\right\|} \right)^{-1}
\end{split}
\end{equation}
From here I get stuck. How should I proceed? Would it makes sense to consider a linear system $Ax = y$ such that $x = A^{-1}y$ and write
$\| cA^{-1} \| = \max_{y\neq 0}\frac{|c| \cdot \left\|A^{-1}y\right\|}{\left\|y\right\|}$ ?
Best Answer
I am writing this answer for the sake of completeness and it will be easier for others to get the answer rather than looking through the comments.
We need to prove that $\kappa{(c * A)} = \kappa{(A)} $.
Consider $$ \kappa{(c * A)} = ||(c * A)|| . ||(c * A)^{-1}|| $$ $$ = |c| . ||A|| . ||c^{-1} A^{-1}|| $$ $$ = |c| . ||A|| . |c^{-1}| . ||A^{-1}|| $$ $$ = |c| . |\frac{1}{c}| . \kappa{(A)}$$ $$ = \kappa{(A)} $$ Hence, proved.