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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

$\| cA^{-1} \| = \max_{y\neq 0}\frac{|c| \cdot \left\|A^{-1}y\right\|}{\left\|y\right\|}$ ?

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  • $\begingroup$ First, $\kappa(cA)=\|cA\|\|(cA)^{-1}\|$. Second, why not to use directly a norm property that $\|cA\|=|c|\|A\|$? $\endgroup$ Commented Nov 10, 2016 at 22:25
  • $\begingroup$ I ovelooked that it should be $\| (cA)^{-1} \|$ instead of $\| cA^{-1} \|$ in $\kappa(cA)$. That would give: $\kappa(cA)=\|cA\|\|(cA)^{-1}\| =|c|\|A\| |c^{-1} \| A^{-1} \| = \|A\| \|A^{-1} \| = \kappa(A)$. $\endgroup$
    – clubkli
    Commented Nov 10, 2016 at 23:30

1 Answer 1

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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.

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