Is the action of taking the inverse distributive in any way? For example, is $(A+B)^{-1}= A^{-1} + B^{-1}?$

Question

Sorry for not formatting properly, I can't seem to get the exponents to show up properly (ex. $A^{-1}$).

Can you distribute the act of taking the inverse over a pair of brackets?

For example, is $(A+B)^{-1} = A^{-1} + B^{-1}?$

That doesn't seem right, but there are some proofs that I can't see being possible without having a place to start, and the only place to start seems to be to do something with the inverse operator (if I can call it an operator). For example: Show that $(C+DD^{T})^{-1}=C^{-1}D(I+D^{T}C^{-1}D)^{-1}$

I've no idea how proofs on that post are getting started; I don't understand their first step. For example:

$(C+DD^T)^{-1} =((I+DD^TC^{-1})C)^{-1}$

Why is the above true, and how did the answerer get there? This is why I'm thinking there's some kind of way to expand an inverse.

It always seems to involve the identity matrix, and I realize that $A^{-1}A=I,\;$ but I'm still unable to understand those proofs and how they're expanding the expression out in the way that they are.

Any help whatsoever is greatly appreciated.

Answer 1

Consider the case of $1\times 1$ matrices.

Observe \begin{align} (1+1)^{-1} \neq 1^{-1}+1^{-1}. \end{align}

Answer 2

If $A=I$ and $B=-I$ are $n\times n$ matrices, both are invertible and equal their own inverses, but their sum is $0$ matrix

Answer 3

Actually you can show it more generally. If you show that matrices with standard matirx-addition do have group structure. and it becomes obvious, as for any group, if a,b are elements of your Group (a*b)^-1 =b^-1 * a^-1.

Answer 4

Even doesn't hold for diagonal case. Consider$$A=diag(a_1,a_2,\cdots, a_n )\\B=diag(b_1,b_2,\cdots, b_n )$$The we have $$(A+B)^{-1}=diag\left({1\over a_1+b_1},{1\over a_2+b_2},\cdots ,{1\over a_n+b_n}\right)$$and $$A^{-1}+B^{-1}=diag\left({1\over a_1}+{1\over b_1},{1\over a_2}+{1\over b_2},\cdots ,{1\over a_n}+{1\over b_n}\right)$$clearly $${1\over a_i+b_i}\ne {1\over a_i}+{1\over b_i}$$