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.

3This [article](http://www.jstor.org/pss/2690437) 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

Related : https://math.stackexchange.com/q/2977195/2987 – Rajesh D Oct 30 '18 at 11:18
12 Answers
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 $(A+B)^{1}=A^{1}$).
 1
 30
 276
 565
 356,881
 50
 750
 1,081

3

26The lemma is the [ShermanMorrison formula](https://en.wikipedia.org/wiki/Sherman%E2%80%93Morrison_formula), isn't it? – Apr 30 '14 at 08:15

2Can 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 fourvectors? Please see this question in Physics.SE: http://physics.stackexchange.com/q/141613/31965 Thanks. – Physics_maths Oct 16 '14 at 15:59

8What about the case of $ \left( A + \lambda I \right)^{1} $? Let's assume $ A $ is PSD. – Royi Aug 22 '17 at 07:44

2I am also interested in the case $(\mathbf{A}+\mathbf{I})^{1}$. Please see https://math.stackexchange.com/questions/2680914/inverseofsymmetricmatrixplusidentitymatrix?noredirect=1#comment5539327_2680914 – TheDon Mar 09 '18 at 16:00


It may be useful to note that the rank of the outer product, $\mathbf{u}\mathbf{v^T}$, of nonzero vectors $\mathbf{u}$ and $\mathbf{v}$, is 1. So the lemma can be used in cases where $B=\mathbf{u}\mathbf{v^T}$, which may come up in linear regression settings when $B = \mathbf{x_i}\mathbf{x_i^T}$ – bob Apr 10 '20 at 16:44

That's a nice lemma. Used it to solve "inverse of a variance in random effects ANOVA" in 3 lines! – Ufos Apr 25 '21 at 14:45

1@bob, every rank one matrix has that form (i.e. is outer product of two vectors). – Peter Morfe Jun 24 '21 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 $B$.
 4,740
 3
 29
 41

11What about the case of $ \left( A + \lambda I \right)^{1} $? Let's assume $ A $ is PSD. – Royi Aug 22 '17 at 07:45

3This follows directly from Woodbury Matrix Identity. Let C=I, V=I. https://en.m.wikipedia.org/wiki/Woodbury_matrix_identity – Mark Borgerding May 22 '18 at 17:28

1
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
 603
 5
 9

3

2How 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

2

6In 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 05:43

2@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. – Ryan Budney 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)
 71,951
 6
 191
 335
 22,222
 3
 64
 104
I'm surprising that no one realize it's a special case of the wellknown 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}$ .
 754
 7
 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 $\epsilon=1$.
 37,360
 2
 50
 99

2The 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 02:20

What about the case of $ \left( A + \lambda I \right)^{1} $? Let's assume $ A $ is PSD. – Royi Aug 22 '17 at 07:45

Royi check Neumann series https://en.wikipedia.org/wiki/Neumann_series – fr_andres May 29 '21 at 05:37
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 lowrank, then you'd be happy using SMW.


The ShermanMorrison "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 lowrank, and is written as $B=UCV^T$. – 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 $k$.
 16,798
 11
 59
 85
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.
 504
 1
 5
 14
 518
 4
 3
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 $IX+X^2+\cdots$ 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.
 24,082
 9
 33
 83
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.
 21
 2
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 $Ax=yBx$, 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: $$(A+B)^{1}=A^{1}(I (I+BA^{1})^{1}BA^{1})$$
 522
 4
 7