# 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. Jan 16 '11 at 20:58
• Oct 30 '18 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 $$(A+B)^{-1}=A^{-1}$$).

• Thanks, I was looking for something like this. Jan 17 '11 at 15:18
• The lemma is the Sherman-Morrison formula, isn't it?
– user856
Apr 30 '14 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. Oct 16 '14 at 15:59
• What about the case of $\left( A + \lambda I \right)^{-1}$? Let's assume $A$ is PSD.
– Royi
Aug 22 '17 at 7:44
• I am also interested in the case $(\mathbf{A}+\mathbf{I})^{-1}$. Please see math.stackexchange.com/questions/2680914/… Mar 9 '18 at 16:00

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
Aug 22 '17 at 7:45
• This follows directly from Woodbury Matrix Identity. Let C=I, V=I. en.m.wikipedia.org/wiki/Woodbury_matrix_identity May 22 '18 at 17:28
• Oct 30 '18 at 11:19

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

• Where did you find this? Can you give a citation? 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 '14 at 15:15
• Interesting to notice that line 3 is a Sylvester equation.
– ati
Dec 11 '14 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
Feb 16 '15 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. Oct 29 '19 at 18:58

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

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)

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 $1$. Jan 17 '11 at 2:20
• What about the case of $\left( A + \lambda I \right)^{-1}$? Let's assume $A$ is PSD.
– Royi
Aug 22 '17 at 7:45
• Royi check Neumann series en.wikipedia.org/wiki/Neumann_series May 29 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? 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. 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

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 $k$.

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: $$\begin{equation} (A+B)^{-1} = A^{-1} + X \end{equation}$$ $$\begin{equation} X = - (I + A^{-1}B)^{-1} A^{-1} B A^{-1} \end{equation}$$ So $$\begin{equation} (A+B)^{-1} = A^{-1} - (I + A^{-1}B)^{-1} A^{-1} B A^{-1} \tag{1}\label{eq1} \end{equation}$$ This rearranges to: $$\begin{equation} (A+B)^{-1} = (I - (I + A^{-1}B)^{-1} A^{-1} B )A^{-1} \tag{2}\label{eq2} \end{equation}$$ If we consider the part $$\begin{equation} (I + A^{-1}B)^{-1} \end{equation}$$ 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: $$\begin{equation} (I + A^{-1}B)^{-1} = (I - (I + A^{-1}B)^{-1}A^{-1}B ) \end{equation}$$ so we can substitute the LHS of this for the right hand side which appears in \eqref{eq2}, giving: $$\begin{equation} (A+B)^{-1} = (I + A^{-1}B)^{-1}A^{-1} \tag{3}\label{eq3} \end{equation}$$ Which is simpler than \eqref{eq1} and is very similar to the scalar identity: $$\begin{equation} \frac{1}{a+b}=\frac{1}{\left(1+\frac{b}{a}\right)a} \tag{4}\label{eq4} \end{equation}$$

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.

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