Intereting Posts

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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.

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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.

$$(A+B)(A^{-1}+B^{-1})=2I+AB^{-1}+BA^{-1}$$

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.

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 $$(A+B)^{-1}=(1+e^{-2\pi i /3})^{-1}I=(1+e^{2\pi i /3})I = A^{-1}+B^{-1}.$$

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?”

- Every statement is unknown until proven true or false.
- Iff a statement is unknown or false, then you cannot use it.
- $\therefore$ 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 $(A+B)^{-1}=[\tfrac 13]$

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:

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

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!

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