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Condition Number of sum of Matrices

Abbas

New member
Aug 9, 2013
4
As far as I know there is no explicit formulas but is this true? I've tested it in Matlab with random matrices and It seems true!
cond(A+B) =< cond(A) + cond(B)
Where can I find a proof for this hypothesis?
 

Abbas

New member
Aug 9, 2013
4
Welcome to MHB, Abbas! :)

\begin{aligned}
\text{cond}(A+B)
&= ||(A+B)^{-1}|| \cdot ||A+B|| \\
&= ||A^{-1} + B^{-1}||\cdot ||A+B|| \\
&\le \Big(||A^{-1}||+||B^{-1}||\Big) \cdot \Big(||A||+||B||\Big) \\
&\le ||A^{-1}||\cdot||A|| + ||B^{-1}||\cdot||B|| \\
&= \text{cond}(A) + \text{cond}(B)
\end{aligned}
Thanks, but Are you sure if this is true?
I doubt (A+B)-1= A-1+B-1.
How about ||A-1||⋅||B||+||A||⋅||B-1|| ? can these terms be omitted?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,855
Thanks, but Are you sure if this is true?
I doubt (A+B)-1= A-1+B-1.
How about ||A-1||⋅||B||+||A||⋅||B-1|| ? can these terms be omitted?
You're quite right. I had just deleted my post, since I realized it was not correct due to the very reasons you mention.
 

Abbas

New member
Aug 9, 2013
4
You're quite right. I had just deleted my post, since I realized it was not correct due to the very reasons you mention.
I was looking for an answer since I post it here, cond(A+B) =< cond(A) + cond(B) is not always true. the hypothesis is wrong. Thanks BTW. :)