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Given a $n \times n$ matrix $A$ and $B$, we need to prove $k(AB) \leq k(A)k(B)$ where $k(\cdot)$ denotes the condition number of a matrix.

Is there any thing wrong in the below proof?

$k(AB) = \|AB\| \cdot \|(AB)^{-1}\| \leq \|A\| \cdot \|B\| \cdot \|B^{-1}\| \cdot \|A^{-1}\| =k(A)k(B) .$

  • 1
    Is $\kappa(AB)\leq\kappa(A)\kappa(B)$ still true when $A$ and $B$ are not both invertible?2013-11-11

2 Answers 2

1

You proof is correct. Both $A, B$ should be invertible, so $\|(AB)^{-1}\|=\|B^{-1}A^{-1}\|\le \|B^{-1}\|\|A^{-1}\| $.

0

$||(AB)||\leq ||A||||B||$ is not true for a general matrix norm. For example, $||A||=max|a_{i,j}|$ and let $A=B=[1,1;1,1]$.