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I feel like I have seen this bound before but no longer can recall its source:

Suppose $A, B$ are square matrices with $||A^{-1} (A-B) || < 1$, i.e. their perturbation is relatively minimal. Then, what is an upper bound on $|| A^{-1} - B^{-1} ||$?

In my notes, I have: $$ || A^{-1} - B^{-1} || \leq \frac{||A^{-1}||^2 ||A - B||}{1 - ||A^{-1} (A - B)||} $$

Is this true? And is there a source?

C. Kang
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2 Answers2

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We have $\|I-A^{-1}B\|<1$, so $$B^{-1}A=[A^{-1}B]^{-1}=\sum_{n=0}^\infty(I-A^{-1}B)^n.$$ So $B^{-1}=\sum_{n=0}^\infty (I-A^{-1}B)^nA^{-1}$.

We have \begin{align*} \|A^{-1}-B^{-1}\| & =\Bigl\|\Bigl[I-\sum_{n=0}^\infty (I-A^{-1}B)^n\Bigr]A^{-1}\Bigr\|=\Bigl\|\Bigl[\sum_{n=1}^\infty (I-A^{-1}B)^n\Bigr]A^{-1}\Bigr\| \\ & \leqslant \|A^{-1}\|\sum_{n=1}^\infty \|I-A^{-1}B\|^n= \|A^{-1}\|\cdot \frac{\|I-A^{-1}B\|}{1-\|I-A^{-1}B\|} \\ & = \|A^{-1}\|\cdot \frac{\|A^{-1}(A-B)\|}{1-\|A^{-1}(A-B)\|}\leqslant \|A^{-1}\|\cdot \frac{\|A^{-1}\|\|A-B\|}{1-\|A^{-1}(A-B)\|}\\ & = \|A^{-1}\|^2\cdot \frac{\|A-B\|}{1-\|A^{-1}(A-B)\|}. \end{align*} This is valid for $A,B$ in any Banach algebra $\mathfrak{A}$, since we only used completeness, triangle inequality, and $\|UV\|\leqslant \|U\|\|V\|$ for any $U,V\in\mathfrak{A}$. In particular it holds for square matrices $M^{n\times n}$ with any matrix norm satisfying the inequality $\|UV\|\leqslant \|U\|\|V\|$.

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    Wonderful! This is awesome, exactly what I was looking for. Thank you!

    (and, as an aside, the underlying technique can be applied for the actual problem I wanted to solve. Again, we have A, B but instead want to study:

    $$ || A B^{-1} - I|| $$ Via the series expansion of the inverse, the expression simplifies into a fraction with some ||A - B || ||A^{-1}|| terms, which is useful because B is a perturbation of A).

    Again, many thanks!

     – C. Kang Jan 04 '24 at 02:06
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It is probably worth noting that just because your first inequality holds, the norm $\|A^{-1} - B^{-1}\|$ need not be bounded. As an example, choose $1\times 1$ matrices with $A=1, B=\varepsilon$ with $0<\varepsilon<1$. Then $$ \| A^{-1} [ A-B] \| = |1-\varepsilon| < 1 $$ but $$ \| A^{-1} - B^{-1} \| = \left|1-\frac{1}{\varepsilon}\right|, $$ which is unbounded as $\varepsilon\to 0$.

This kind of argument is easily generalized to larger matrices.

Wolfgang Bangerth
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