# Inverse of the sum of matrices

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.

• @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. Commented Jan 16, 2011 at 20:58
• Commented Oct 30, 2018 at 11:18

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 $$g=\operatorname{trace}(BA^{-1})$$. 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 $$g_k = \frac{1}{1 + \operatorname{trace}(C_k^{-1}B_k)}$$. 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 $$(A+B)^{-1}=A^{-1}$$).

• The lemma is the Sherman-Morrison formula, isn't it?
– user856
Commented Apr 30, 2014 at 8:15
• Can this theorem be used in finding the inverse of $${\large[}g_{\mu\nu}+\chi \frac{k_\mu k_\nu}{k^2}{\large]}$$ where $g$ is the Minkowski metric tensor and the $k$'s are four-vectors? Please see this question in Physics.SE: physics.stackexchange.com/q/141613/31965 Thanks. Commented Oct 16, 2014 at 15:59
• What about the case of $\left( A + \lambda I \right)^{-1}$? Let's assume $A$ is PSD.
– Royi
Commented Aug 22, 2017 at 7:44
• I am also interested in the case $(\mathbf{A}+\mathbf{I})^{-1}$. Please see math.stackexchange.com/questions/2680914/… Commented Mar 9, 2018 at 16:00
• @bob, every rank one matrix has that form (i.e. is outer product of two vectors).
– user711689
Commented Jun 24, 2021 at 16:43

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 $$B$$.

• What about the case of $\left( A + \lambda I \right)^{-1}$? Let's assume $A$ is PSD.
– Royi
Commented Aug 22, 2017 at 7:45
• This follows directly from Woodbury Matrix Identity. Let C=I, V=I. en.m.wikipedia.org/wiki/Woodbury_matrix_identity Commented May 22, 2018 at 17:28
• Commented Oct 30, 2018 at 11:19

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

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)

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

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

• Where did you find this? Can you give a citation? Commented Jun 11, 2013 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
Commented Apr 10, 2014 at 15:15
• Interesting to notice that line 3 is a Sylvester equation.
– ati
Commented Dec 11, 2014 at 16:26
• In order to conclude last line,we must have (I+A^-1B) invertible. So how are we sure about that, It might be easy but (I am not getting. Can you please explain @ Muhammad Fuday.
– Sry
Commented Feb 16, 2015 at 5:43
• @Sry: I'm not certain how this formula helps. For example, the deduction $(I+A^{-1}B)^{-1} = (A+B)^{-1} A$ is direct, so the above formula is basically just the statement $(A+B)^{-1})(A+B)=I$. Among other things $I+A^{-1}B$ is invertible if and only if $A+B$ is invertible. i.e. you have to check invertibility of two equivalent matrices. Commented Oct 29, 2019 at 18:58

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

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

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 $\epsilon=1$.

• 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 $1$. Commented Jan 17, 2011 at 2:20
• What about the case of $\left( A + \lambda I \right)^{-1}$? Let's assume $A$ is PSD.
– Royi
Commented Aug 22, 2017 at 7:45
• Royi check Neumann series en.wikipedia.org/wiki/Neumann_series Commented May 29, 2021 at 5:37

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.

• It also requires $(A^{-1} + B^{-1})^{-1}$ to be known, doesn't it? Commented Jan 16, 2011 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. Commented Jan 16, 2011 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
Commented Jan 16, 2011 at 22:01

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 $k$.

I know the question has been answered multiple times with great answers, but with my answer you don't need to memorize any lemmas or formulas.

Suppose $$(A+B)x=y$$, then $$x=(A+B)^{-1}y$$. This is all we need to get. The steps are:

(1) Start with $$(A+B)x=y$$.

(2) Then $$Ax=y-Bx$$, so $$x=A^{-1}y -A^{-1}Bx$$.

(3) Multiply $$x$$ in step (2) by $$B$$ to get $$Bx=BA^{-1}y -BA^{-1}Bx$$ which is equivalent to $$(I+BA^{-1})Bx=BA^{-1}y$$ or, $$Bx=(I+BA^{-1})^{-1}BA^{-1}y$$

(3) Substitute this $$Bx$$ into the $$x$$ in step (2) to get $$x=A^{-1}y -A^{-1}(I+BA^{-1})^{-1}BA^{-1}y$$

(4) Now factorizing the $$y$$ gives you the required result. $$x=(A^{-1} -A^{-1}(I+BA^{-1})^{-1}BA^{-1})y$$

(5)The assumptions we have used are $$A$$ and $$I+BA^{-1}$$ are nonsingular.

(6) We can factorize the $$A^{-1}$$ to get: $$(A+B)^{-1}=A^{-1}(I -(I+BA^{-1})^{-1}BA^{-1})$$

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 $$A^{-1}$$ nor $$B^{-1}$$ exist but $$(A+B)^{-1}$$ does or, conversely, such that both $$A^{-1}$$ and $$B^{-1}$$ exist but $$(A+B)^{-1}$$ doesn't.

Extending Muhammad Fuady's approach: We have: $$$$(A+B)^{-1} = A^{-1} + X$$$$ $$$$X = - (I + A^{-1}B)^{-1} A^{-1} B A^{-1}$$$$ So $$$$(A+B)^{-1} = A^{-1} - (I + A^{-1}B)^{-1} A^{-1} B A^{-1} \tag{1}\label{eq1}$$$$ This rearranges to: $$$$(A+B)^{-1} = (I - (I + A^{-1}B)^{-1} A^{-1} B )A^{-1} \tag{2}\label{eq2}$$$$ If we consider the part $$$$(I + A^{-1}B)^{-1}$$$$ Then, this is an inverse of a sum of two matrices, so we can use \eqref{eq2}, setting $$A=I$$ and $$B = A^{-1}B$$, this gives: $$$$(I + A^{-1}B)^{-1} = (I - (I + A^{-1}B)^{-1}A^{-1}B )$$$$ so we can substitute the LHS of this for the right hand side which appears in \eqref{eq2}, giving: $$$$(A+B)^{-1} = (I + A^{-1}B)^{-1}A^{-1} \tag{3}\label{eq3}$$$$ Which is simpler than \eqref{eq1} and is very similar to the scalar identity: $$$$\frac{1}{a+b}=\frac{1}{\left(1+\frac{b}{a}\right)a} \tag{4}\label{eq4}$$$$

The technique is useful in computation, because if the values in A and B can be very different in size then calculating $$\frac{1}{A+B}$$ according to \eqref{eq3} gives a more accurate floating point result than if the two matrices are summed.

Actually we can directly from @Shiyu answer about perturbations by subtracting $$(A+B)^{-1}$$ and factoring arrive at

$$0=A^{-1}-(A^{-1}B+I)(A+B)^{-1}$$ followed by$$(A+B)^{-1}=(A^{-1}B+I)^{-1}A^{-1}$$

And by symmetry of course

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

Now remember, $$(I+X)^{-1}$$ can be expanded as $$I-X+X^2+\cdots$$ by geometric series.

So if $$X=B^{-1}A$$ or $$X=A^{-1}B$$ and multiplication by $$A,B$$ and either of $$A^{-1}$$ or $$B^{-1}$$ are cheap, then this could work nicer than some other method of finding inverse.

By means of augmented matrix,

A+B|I

Left times A⁻¹:

I+A⁻¹B|A^-1

abstract the common factor B:

(B^-1+A^-1)B|A^-1

left times (B^-1+A^-1)^-1:

B|(B^-1+A^-1)⁻¹A^-1

left times B^-1:

I|B^-1(B^-1+A^-1)⁻¹A^-1

thus (A+B)⁻¹=B⁻¹(B⁻¹+A⁻¹)⁻¹A⁻¹

• We have a Latex-like typesetting system for mathematical expressions, called MathJax. Information is here: math.meta.stackexchange.com/a/10164 Commented Oct 24, 2022 at 13:24