Upper Bound for the Induced 2-Norm $\|(A+B)^{-1}A\|$

Question

Assume $A$ is a positive definite matrix, and $B$ is a positive semi-definite matrix. I am interested in the problem of whether there exists a constant upper bound for the induced 2-norm (spectral norm) of $\|(A+B)^{-1}A\|$.

I have tried that $\|(A+B)^{-1}A\|\leq \|(A+B)^{-1}\|\|A\|=\frac{\max eigenvalue(A)}{\min eigenvalue(A+B)}\leq \frac{\max eigenvalue(A)}{\min eigenvalue(A)} $.

However, the above upper bound depends on the condition number of a specific matrix $A$. I wonder if there is any universal constant upper bound for the induce 2-norm $\|(A+B)^{-1}A\|$, such as 10, 2, or $\sqrt{2}$?

Thank you!

Answer 1

No, $\|(A+B)^{-1}A\|_2$ is not bounded above. Pick any real upper triangular matrix $C$ whose diagonal entries lie inside $(0,1]$. Modify its strictly upper triangular part to make $\|C\|_2$ arbitrarily large. By construction, $C^{-1}-I$ has a real nonnegative spectrum. Hence (cf. Olga Taussky, problem 4846, Amer. Math. Monthly 66, 1959, p.427) $C^{-1}-I=PS$ for some positive definite matrix $P$ and positive semidefinite matrix $S$. By putting $A=P^{-1}$ and $B=S$, we obtain an example pair $(A,B)$ such that $(A+B)^{-1}A=(I+A^{-1}B)^{-1}=C$ has an arbitrarily large norm.

For a concrete example, let $m>0$ be a very large number, $$ A^{-1}=\pmatrix{1&-(6m)^{-1}\\ -(6m)^{-1}&(18m^2)^{-1}} \text{ and } B=\pmatrix{1&3m\\ 3m&9m^2}. $$ One can easily verify by Sylvester's criterion that $A^{-1}\succ0$ and $B\succeq0$. Now \begin{aligned} I+A^{-1}B &=I+\pmatrix{1&-(6m)^{-1}\\ -(6m)^{-1}&(18m^2)^{-1}}\pmatrix{1&3m\\ 3m&9m^2}\\ &=I+\pmatrix{1/2&3m/2\\ 0&0}=\pmatrix{3/2&3m/2\\ 0&1}.\\ \end{aligned} Therefore $$ (A+B)^{-1}A=(I+A^{-1}B)^{-1}=\pmatrix{2/3&-m\\ 0&1}, $$ whose induced $2$-norm is greater than $m$.