5
$\begingroup$

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?

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

  • 3
    In the third line, one of the $B^{-1}$ should be $A^{-1}$, otherwise it looks fine. If this is homework you might also be required to show that $\| A B \| \leq \|A \| \|B\|$, unless you did this in your class/lecture.2011-11-02
  • 2
    If you want the Greek letter $\kappa$ instead of $k$ for the condition number, then use `$\kappa$`.2011-11-02
  • 1
    As @Oliver pointed out, your reasoning is correct. One important point is to note that *submultiplicative* property of the norm used. When matrices are seen as operators, not all operator norms are submultiplicative. But for *condition number*, usually Euclidean norm is taken and it is submultiplicative.2011-11-02
  • 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 $A=B=[1,1;1,1]$.