Knowing that K(A)= ||A||.||A^-1||, also known as the condition number A is n by n matrix and c is element of real numbers that is greater than zero.
Show that 1) K(AB)<= K(A)*K(B) 2) c.K(AB) =K(AB)
Question about condition numbers properties
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linear-algebra
number-theory
numerical-methods
proof-explanation
gaussian-elimination
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1I think to show 1, you need to use the fact that $\|AB\| \leq \|A\|\|B\|$.For instance, we have: $$K(AB) = \|AB\| \|A^{-1}B^{-1}\| \leq \|A\|\|B\|\|A^{-1}B^{-1}\| \leq\|A\|\|B\|\|A^{-1}\|\|B^{-1}\| = (\|A\|\|A^{-1}\|)(\|B\|\|B^{-1}\|) = K(A)K(B),$$ which is the required result. – 2017-02-20
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0It's not clear what 2) is *supposed* to say. I think it should be $$ \kappa(c \cdot A) = \kappa(A) $$ which can be prove using the properties of the norm – 2017-02-20
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0Thank you for your reply. I got something similar for part a, but part B is saying if we have a scaler inside the condition number, the condition number will stay the same, but I don'5 know – 2017-02-20