Let $A, B,C$ be invertible matrices, and $$d(A,B)\ :=\ 1-\frac{1}{\mbox{cond}(A^{-1}B)},$$ where $\mbox{cond}(E)=\|E\|\cdot\|E^{-1}\|$ and $\|\cdot\|$ a subordinate matrix norm. Somebody knos how do I can prove that $$d(A,C)\ \leq\ d(A,B)+d(B,C)?$$ Thanks for help in advance.


1 Answer 1


(Presumably the term "matrix norm" refers to a submultiplicative matrix norm, i.e. one that satisfies $\|AB\|\le\|A\|\|B\|$.)


  1. Show that for any square matrix $X$, we have $\operatorname{cond}(X)\ge1$.
  2. Let $P=A^{-1}B$ and $Q=B^{-1}C$. Show that the target inequality is equivalent to $$ 1-\frac1{\operatorname{cond}(P)}-\frac1{\operatorname{cond}(Q)}+\frac1{\operatorname{cond}(PQ)}\ge0,\\ $$
  3. By splitting fractions of the form $\frac{\|PQx\|}{\|x\|}$ into the product $\frac{\|P(Qx)\|}{\|Qx\|}\frac{\|Qx\|}{\|x\|}$, show that $\operatorname{cond}(PQ)\le\operatorname{cond}(P)\operatorname{cond}(Q)$.
  4. Use results from steps 1-3 to show that the target inequality is true.
  • $\begingroup$ Thanks so much, I get it!! $\endgroup$
    – FASCH
    Apr 14, 2013 at 15:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .