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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.

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    Related : https://math.stackexchange.com/q/2977195/29872018-10-30

12 Answers 12

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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 $(A+B)^{-1}=A^{-1}$).

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It is shown in On Deriving the Inverse of a Sum of Matrices that

$(A+B)^{-1}=A^{-1}-A^{-1}B(A+B)^{-1}$.

This equation cannot be used to calculate $(A+B)^{-1}$, but it is useful for perturbation analysis where $B$ is a perturbation of $A$. There are several other variations of the above form (see equations (22)-(26) in this paper).

This result is good because it only requires $A$ and $A+B$ to be nonsingular. As a comparison, the SMW identity or Ken Miller's paper (as mentioned in the other answers) requires some nonsingualrity or rank conditions of $B$.

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    Related : https://math.stackexchange.com/q/2977195/29872018-10-30
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$(A+B)^{-1} = A^{-1} - A^{-1}BA^{-1} + A^{-1}BA^{-1}BA^{-1} - A^{-1}BA^{-1}BA^{-1}BA^{-1} + \cdots$

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)

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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

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    What about the case of $ \left( A + \lambda I \right)^{-1} $? Let's assume $ A $ is PSD.2017-08-22
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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 $\epsilon=1$.

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    What about the case of $ \left( A + \lambda I \right)^{-1} $? Let's assume $ A $ is PSD.2017-08-22
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I'm surprising that no one realize it's a special case of the well-known matrix inverse lemma or [Woodbury matrix identity], it says,

$ \left(A+UCV \right)^{-1} = A^{-1} - A^{-1}U \left(C^{-1}+VA^{-1}U \right)^{-1} VA^{-1}$ ,

just set U=V=I, it immediately gets

$ \left(A+C \right)^{-1} = A^{-1} - A^{-1} \left(C^{-1}+A^{-1} \right)^{-1} A^{-1}$ .

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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.

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    @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 $B=UCV^T$.2011-01-16
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It is possible to come up with pretty simple examples where $A$,$A^{-1}$,$B$, and $B^{-1}$ are all very nice, but applying $(A+B)^{-1}$ is considered very difficult.

The canonical example is where $A = \Delta$ is a finite difference implementation of the Laplacian on a regular grid (with, for example, Dirichlet boundary conditions), and $B=k^2I$ is a multiple of the identity. The finite difference laplacian and it's inverse are very nice and easy to deal with, as is the identity matrix. However, the combination $\Delta + k^2 I$ is the Helmholtz operator, which is widely known as being extremely difficult to solve for large $k$.

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If A and B were numbers, there is no simpler way to write $\frac{1}{A+B}$ in term of $ \frac{1}{A}$ and $B$ so I don't know why you would expect there to be for matrices. It is even possible to have matrices, A and B, so that neither $A^{-1}$ and $B^{-1}$ exist but $(A+B)^{-1}$ does or, conversely, such that both $A^{-1}$ nor $B^{-1}$ exist but $(A+B)^{-1}$ doesn't.