# Prove $k(AB) \leq k(A)k(B)$ where $k(\cdot)$ denotes the condition number

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?

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

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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. –  Oliver Braun Nov 2 '11 at 5:38
If you want the Greek letter $\kappa$ instead of $k$ for the condition number, then use $\kappa$. –  Srivatsan Nov 2 '11 at 5:41
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. –  Tapu Nov 2 '11 at 5:59
Is $\kappa(AB)\leq\kappa(A)\kappa(B)$ still true when $A$ and $B$ are not both invertible? –  Randel Nov 11 '13 at 22:20

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