For $A, B$ invertible matrices of the same order, is it true that $(A^T + B^T)^{-1} = (A^T)^{-1} + (B^T)^{-1}$?

Question

If $A$ and $B$ are invertible matrices of the same order then is this statement true? Why? $$(A^T + B^T)^{-1} = (A^T)^{-1} + (B^T)^{-1}$$

Answer 1

No.

Multiplying both sides of that equation by $A^T + B^T$ gives $$I = (A^T + B^T)((A^T)^{-1} + (B^T)^{-1}) = 2I + B^T (A^T)^{-1} + A^T (B^T)^{-1},$$ which reduces to $$A^T (B^T)^{-1} + B^T (A^T)^{-1} = -I$$ and that looks pretty dubious.

For an easy counterexample, consider $A=2I$ and $B = I$. Then $A=A^T$, $A^{-1} = \frac12 I$ and $B=B^T = B^{-1}$. So we have $$A^T + B^T = 2I + I = 3I \qquad (A^T)^{-1} + (B^T)^{-1} = \tfrac12I + I = \tfrac32I,$$ and we see that the original statement fails.