Suppose $A + B$ is invertible, then is it true that $(A + B)^{-1} = A^{-1} + B^{-1}$?

I know the answer is no, but don't get why.

  • 73
    $\begingroup$ This is not true even for numbers, why should it be true for matrices? For example $(1+1)^{-1} \neq 1^{-1} + 1^{-1} = 1+1$ $\endgroup$
    – Crostul
    Commented Feb 17, 2015 at 23:38
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    $\begingroup$ Why is 4 not equal to 5? $\endgroup$ Commented Feb 18, 2015 at 16:23
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    $\begingroup$ Math teachers complain how hard it is to get students to stop assuming that things like this would work. There are transformations (such as multiplication by a scalar) that distribute over addition in suitable domains (such as the real numbers), but it's actually interesting and remarkable to show that something does distribute over addition; the assumption should be that it does not. $\endgroup$
    – David K
    Commented Feb 18, 2015 at 18:06
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    $\begingroup$ When people ask “Why isn’t (such-and-such) true?” I like to answer: In mathematics, nothing is true, unless you have a proof for it. $\endgroup$
    – Lubin
    Commented Feb 18, 2015 at 22:56
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    $\begingroup$ I don't understand, when people have such a question, why don't they just try any random example. In this case it is very hard to come up with an example for which the equality is true. $\endgroup$ Commented Feb 19, 2015 at 20:29

8 Answers 8


Let us look at the often-forgotten $1 \times 1$ matrices over $\mathbb{R}$, which is another name for the real numbers themselves. Then your statement translates to the statement that if $x,y$ are real numbers that aren't zero, then $$ \frac{1}{x + y} = \frac{1}{x} + \frac{1}{y},$$ which is clearly wrong.

  • 15
    $\begingroup$ Great answer. +1 $\endgroup$ Commented Feb 17, 2015 at 23:39
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    $\begingroup$ Just a fun remark: for the even more often forgotten $0\times0$ matrices, the equation of the question does hold. But then for such matrices (the plural really is out of place) any equation holds (including Cayley-Hamilton, which states $I=0$). $\endgroup$ Commented Feb 18, 2015 at 17:32
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    $\begingroup$ @asmeurer The reals are (isomorphic to) a subring of the ring of $n\times n$ matrices (for $n>0$, to keep Marc van Leeuwen happy ;-)), so if an identity holds for matrices it must hold for the reals. $\endgroup$
    – egreg
    Commented Feb 19, 2015 at 11:45
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    $\begingroup$ @PeterLeFanuLumsdaine: Indeed; $\det(0I_n)=0^n$. $\endgroup$ Commented Feb 20, 2015 at 5:25
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    $\begingroup$ @ mixedmath , indeed the extraordinary subtlety of your answer very much deserves 111 (or more) upvotes ! $\endgroup$
    – user91684
    Commented Feb 20, 2015 at 15:45

$$(A+B)(A^{-1}+B^{-1})=2I+AB^{-1}+BA^{-1}$$ so your statement is true if and only if $I+AB^{-1}+BA^{-1}=0$ (which is of course not always true, and even usually wrong).


To top off the list of examples, take $A=I$ and $B=-I$. Then the two matrices are invertible but their sum is not.

  • $\begingroup$ this was the example my book gave. so why is the sum not invertible? $\endgroup$
    – fatwalrus
    Commented Feb 17, 2015 at 23:43
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    $\begingroup$ Because it is zero. $\endgroup$ Commented Feb 17, 2015 at 23:44
  • $\begingroup$ Just dropping by and I don't khow much, but isn't zero the inverse of itself? $\endgroup$
    – Ooker
    Commented Feb 19, 2015 at 12:44
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    $\begingroup$ Additive yes, multiplicative no $\endgroup$ Commented Feb 19, 2015 at 13:19
  • $\begingroup$ "Because it is zero." The powerful comment of nil :P $\endgroup$
    – Mr Pie
    Commented Apr 8, 2019 at 16:46

Take for example $A= I$ (the identity) and $B = 0$ (the matrix with $0$ everywhere). Then $A+B$ is invertible however $B^{-1}$ does not exists and so the statement makes no sense.

NOTE @JeppeStigNielsen asks in the comment if we can find matrices for which the equality is true. Here is a way to build such a pair. From @anderstood's answer, we know that we have to find $A,B$ such that $AB^{-1}+BA^{-1}+I=0$. Let $C= AB^{-1}$, then this equation can be rewritten as $C+C^{-1}+I = 0$. Suppose that $C = \gamma I$ for some $\gamma \in \Bbb C\setminus \{0\}$.Then we get the equation $\gamma^2+\gamma+1=0$, and so $\gamma \in \{e^{2\pi i/3},e^{4\pi i /3}\}$. Now it stays to find invertible matrices $A,B$ such that $A^{-1}B = \gamma$ and $B^{-1}A=\gamma^{-1}$. Again, let us assume that $A = \alpha I$ and $B = \beta I$ for some $\alpha,\beta \in \Bbb C\setminus \{0\}$. Thus, we need $\alpha\beta^{-1} = \gamma$ and $\beta \alpha^{-1}=\gamma^{-1}$ (note that these equations are equivalent). So we may choose $\alpha \in \Bbb C\setminus \{0\}$ and set $\beta = \alpha\gamma^{-1}$. This gives a whole family of matrices for which the equation is true. Example: $A = I,B=e^{-2\pi i/3}I$, then $$(A+B)^{-1}=(1+e^{-2\pi i /3})^{-1}I=(1+e^{2\pi i /3})I = A^{-1}+B^{-1}.$$

  • $\begingroup$ While finding a counterexample to the statement of the question is extremely easy, it is quite hard to find an example where $(A+B)^{-1}=A^{-1}+B^{-1}$ is actually true. $\endgroup$ Commented Feb 18, 2015 at 9:34
  • $\begingroup$ @JeppeStigNielsen I edited my post to give you a family of examples for which the equation holds. $\endgroup$
    – Surb
    Commented Feb 18, 2015 at 10:05
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    $\begingroup$ I see it. And by using a matrix model (over $\mathbb{R}$) of the complex numbers, this also gives examples of real matrices that solve this equation. $\endgroup$ Commented Feb 18, 2015 at 10:12
  • $\begingroup$ @Surb: I think you made a typo: the condition is that $AB^{-1}+BA^{-1}+I=0$ (you wrote $A^{-1}B$). And if $C=AB^{-1}$, then $C^{-1}=BA^{-1}$ so indeed $C+C^{-1}+I=0$. $\endgroup$
    – anderstood
    Commented Feb 18, 2015 at 18:03
  • $\begingroup$ @anderstood right.. thanks $\endgroup$
    – Surb
    Commented Feb 18, 2015 at 18:28

To answer your question directly $$(A+B)^{-1} \neq A^{-1} + B^{-1}$$ because $$([1] + [1])^{-1} \neq [1]^{-1} + [1]^{-1}.$$

To answer the slightly different question of "Why can't I do that?"

  1. Every statement is unknown until proven true or false.
  2. Iff a statement is unknown or false, then you cannot use it.
  3. $\therefore$ Every statement is unusable until proven true.

Thus in general it is better to ask why may you do something, rather than to ask why may you not do that. So the reason you may not replace $(A+B)^{-1}$ with $A^{-1} + B^{-1}$ is because you has not proven $(A+B)^{-1} = A^{-1} + B^{-1}$ is true (nor can it be proven true because it is false).

So my advice is, do not try to memorize that this statement is false, rather treat it as unknown and thus unusable; because there are too many useless false statements to memorize, so it is best to focus on memorizing the true statements

Now as to why [my proof of it is false] is correct. The statement $(A+B)^{-1} = A^{-1} + B^{-1}$ is really the universal statement $$\forall A, \forall B, (A+B)^{-1} = A^{-1} + B^{-1}.$$ To disprove this universal statement is the same as proving this existential statement $$\exists A, \exists B, (A+B)^{-1} \neq A^{-1} + B^{-1}.$$ Finally to prove a existential statement, one just has to give an example, which I have done.


Very strange... 'I know the answer is no, but don't get why.' Why don't you calculate a sum of two matrices, say size 1 by 1, and compare their inverses to see, why...?

For example $A=[1]$ and $B=[2]$ makes $A^{-1}=[1],\ B^{-1}=[\tfrac 12]$ and $(A+B)^{-1}=[\tfrac 13]$

  • $\begingroup$ I don't think this is an explaination but just one example. An example helps you understand the explaination, but doesn't make one. $\endgroup$
    – Ooker
    Commented Feb 19, 2015 at 12:48
  • $\begingroup$ The explanation is: 'because in general they are not equal, period.' Specifically 'the inverse of a sum is, in general, not equal to the sum of terms' inverses.' If some different expressions turn out to be equivalent, then we seek some specific reasoning, explaining 'why' they are equivalent. But when two different expressions are different, the answer is straightforward: they are different because they are different. Maybe there are some specific cases when they become equal (say, $A=B=\Bbb 1$), but generaly being not equal has no special causes. $\endgroup$
    – CiaPan
    Commented Feb 19, 2015 at 16:31
  • $\begingroup$ @Ooker sometimes answers don't need to be full explanations ;) $\endgroup$
    – Mr Pie
    Commented Apr 8, 2019 at 16:46

Only for completeness, there are matrices $A$ and $B$ such as $$(A+B)^{-1}=A^{-1}+B^{-1},$$
take, for instance,
$$A=\begin{bmatrix} -\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{bmatrix}$$ and B the identity matrix. It is possible to build examples $A_n$ and $B_n$ when the dimension $n$ is even, just take $A_n$ as the diagonal block matrix with $A$ in the diagonal entries and $B_n$ the identity matrix. But there are no examples in odd dimensions!


It's just that the operation of taking inverses isn't well-behaved when it comes to sums.

This isn't surprising, because taking inverses is defined in terms of multiplication and NOT in terms of addition. There is no reason really why inverses should act in some favorable way on sums.

For instance, if $A$ is some invertible matrix, then what is the inverse of $A+A$? Is it $A^{-1}+A^{-1}$? No. You have that $A+A=2A$, whose inverse is $1/2 A^{-1}$, and this is not the same as $A^{-1}+A^{-1}=2A^{-1}$.

What's more is that if $A$ and $B$ are invertible matrices, then their sum $A+B$ need not even be invertible, as others have pointed out: It is not even always possible to find $(A+B)^{-1}$, even though both $A^{-1}$ and $B^{-1}$ exist.

What does hold, which you probably already know, is that (for invertible $A$ and $B$), we have this nice multiplicative property:

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


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