Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have two square matrices - $A$ and $B$. $A^{-1}$ is known and I want to calculate $(A+B)^{-1}$. Are there theorems that help with calculating the inverse of the sum of matrices? In general case $B^{-1}$ is not known, but if it is necessary then it can be assumed that $B^{-1}$ is also known.

share|improve this question
1  
This article may be of help, although I don't have access to it right now. –  Adrián Barquero Jan 16 '11 at 20:54
    
@Arturo: I know that they might not be invertible, but let's assume they are. @Adrian: Unfortunately I don't have direct access to jstor. –  Tomek Tarczynski Jan 16 '11 at 20:58

5 Answers 5

up vote 22 down vote accepted

In general, $A+B$ need not be invertible, even when $A$ and $B$ are. But one might ask whether you can have a formula under the additional assumption that $A+B$ is invertible.

As noted by Adrián Barquero, there is a paper by Ken Miller published in the Mathematics Magazine in 1981 that addresses this.

He proves the following:

Lemma. If $A$ and $A+B$ are invertible, and $B$ has rank $1$, then let $g=\mathrm{trace}(BA^{-1})$. Then $g\neq -1$ and $$(A+B)^{-1} = A^{-1} - \frac{1}{1+g}A^{-1}BA^{-1}.$$

From this lemma, we can take a general $A+B$ that is invertible and write it as $A+B = A + B_1+B_2+\cdots+B_r$, where $B_i$ each have rank $1$ and such that each $A+B_1+\cdots+B_k$ is invertible (such a decomposition always exists if $A+B$ is invertible and $\mathrm{rank}(B)=r$). Then you get:

Theorem. Let $A$ and $A+B$ be nonsingular matrices, and let $B$ have rank $r\gt 0$. Let $B=B_1+\cdots+B_r$, where each $B_i$ has rank $1$, and each $C_{k+1} = A+B_1+\cdots+B_k$ is nonsingular. Setting $C_1 = A$, then $$C_{k+1}^{-1} = C_{k}^{-1} - g_kC_k^{-1}B_kC_k^{-1}$$ where $g_k = \frac{1}{1 + \mathrm{trace}(C_k^{-1}B_k)}$. In particular, $$(A+B)^{-1} = C_r^{-1} - g_rC_r^{-1}B_rC_r^{-1}.$$

(If the rank of $B$ is $0$, then $B=0$, so $(A+B)^{-1}=A^{-1}$).

share|improve this answer
    
Thanks, I was looking for something like this. –  Tomek Tarczynski Jan 17 '11 at 15:18
    
The lemma is the Sherman-Morrison formula, isn't it? –  Rahul Apr 30 at 8:15

$(A+B)^{-1} = A^{-1} - A^{-1}BA^{-1} + A^{-1}BA^{-1}BA^{-1} - A^{-1}BA^{-1}BA^{-1}BA^{-1} + \cdots$

provided $\|A^{-1}B\|<1$ or $\|BA^{-1}\| < 1$ (here $\|\cdot\|$ means norm). This is just the Taylor expansion of the inversion function together with basic information on convergence.

(posted essentially at the same time as mjqxxx)

share|improve this answer

This I found accidentally.

Suppose given $A$, and $B$, where $A$ and $A+B$ are invertible. Now we want to know the expression of $(A+B)^{-1}$ without imposing the all inverse. Now we follow the intuition like this. Suppose that we can express $(A+B)^{-1} = A^{-1} + X$, next we will present simple straight forward method to compute $X$ \begin{equation} (A+B)^{-1} = A^{-1} + X \end{equation} \begin{equation} (A^{-1} + X) (A + B) = I \end{equation} \begin{equation} A^{-1} A + X A + A^{-1} B + X B = I \end{equation} \begin{equation} X(A + B) = - A^{-1} B \end{equation} \begin{equation} X = - A^{-1} B ( A + B)^{-1} \end{equation} \begin{equation} X = - A^{-1} B (A^{-1} + X) \end{equation} \begin{equation} (I + A^{-1}B) X = - A^{-1} B A^{-1} \end{equation} \begin{equation} X = - (I + A^{-1}B)^{-1} A^{-1} B A^{-1} \end{equation}

This lemma is simplification of lemma presented by Ken Miller, 1981

share|improve this answer
1  
Where did you find this? Can you give a citation? –  Daniel Renshaw Jun 11 '13 at 10:55
    
How is this a simplification of the lemma shown in Ken Miller 1981? Are we talking about "On the Inverse of the Sum of Matrices" or any other work? (In any case, I find this property quite useful, just need to cite it properly). –  Rufo Apr 10 at 15:15

A formal power series expansion is possible: $$ \begin{eqnarray} (A + \epsilon B)^{-1} &=& \left(A \left(I + \epsilon A^{-1}B\right)\right)^{-1} \\ &=& \left(I + \epsilon A^{-1}B\right)^{-1} A^{-1} \\ &=& \left(I - \epsilon A^{-1}B + \epsilon^2 A^{-1}BA^{-1}B - ...\right) A^{-1} \\ &=& A^{-1} - \epsilon A^{-1} B A^{-1} + \epsilon^2 A^{-1} B A^{-1} B A^{-1} - ... \end{eqnarray} $$ Under appropriate conditions on the eigenvalues of $A$ and $B$ (such that $A$ is sufficiently "large" compared to $B$), this will converge to the correct result at $\epsilon=1$.

share|improve this answer
    
The point about eigenvalues is apt, because this works even if $\|A^{-1}B\|\geq1$ and $\|BA{^-1}\|\geq1$ as long as the spectral radius of $A^{-1}B$ or $BA^{-1}$ is less than $1$. –  Jonas Meyer Jan 17 '11 at 2:20

Assuming everything is nicely invertible, you are probably looking for the SMW identity (which, i think, can also be generalized to pseudoinverses if needed)

Please see caveat in the comments below; in general if $B$ is low-rank, then you'd be happy using SMW.

share|improve this answer
    
It also requires $(A^{-1} + B^{-1})^{-1}$ to be known, doesn't it? –  mjqxxxx Jan 16 '11 at 21:06
    
The Sherman-Morrison "update" formula is most efficient if $B$ is of low rank. So the usual application (rank one or two if symmetry is to be preserved) doesn't require $B^{-1}$ to exist. –  hardmath Jan 16 '11 at 21:07
    
@mjqxxxx: yes, actually smw does require that inverse, which actually renders this answer useless, unless one is looking for inverses where $B$ is low-rank, and is written as $B=UCV^T$. –  user1709 Jan 16 '11 at 22:01

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.