Linear Algebra, add these two matrices

Let $n$ be a natural number $\geq 2$ and $A$ a matrix $\in M_{n}(K)$. We suppose the matrices $A$ and $I_{n}+A$ are invertible.

Calculate: $(I_{n}+A^{-1})^{-1}+(I_{n}+A)^{-1}$

I know the result is $I_{n}$ but i don't understand why. Here's my effort:

$(I_{n}+A^{-1})^{-1}+(I_{n}+A)^{-1}=(A^{-1})^{-1}+(A)^{-1}=(A)+A^{-1}$

I'm assuming $I_{n}+A=A$ ?

Where did i go wrong. Any help would be greatly appreciated.

Thank you.

Solution 1:

Since $(I+A^{-1})=A^{-1}(I+A)$, we have

$$(I+A^{-1})^{-1}+(I+A)^{-1}=(A^{-1}(I+A))^{-1}+(I+A)^{-1}=(I+A)^{-1}\left(\left(A^{-1}\right)^{-1}+I\right)=(I+A)^{-1}(I+A)=I,$$

where at the last stage we used that $(A^{-1})^{-1}=A$.

Solution 2:

Since this is the accepted answer, I feel obliged to give a short, quick and effective proof which holds in full generality. This is:

$$(I+A^{-1})^{-1}+(I+A)^{-1}=A\cdot A^{-1}(I+A^{-1})^{-1}+(I+A)^{-1}$$ $$=A\cdot \bigg((I+A^{-1})A\bigg)^{-1}+(I+A)^{-1}=A\cdot (A+I)^{-1}+(A+I)^{-1} $$ $$=(A+I)(A+I)^{-1}=I.$$

But I also think there is some value to the original proof I gave (although it only holds for real and complex matrices). Therefore, I leave it below.

First, consider the sets $\mathcal{A}:=\{A \in M_{n \times n}(\mathbb{C}) \mid \exists A^{-1}\}$. and $\mathcal{B}:=\{B \in M_{n \times n}(\mathbb{C}) \mid \exists (I+B)^{-1}\}$. Those are open sets. Put $\mathcal{N}:=\mathcal{A} \cap \mathcal{B}$ and consider the function

$$f: \mathcal{N} \rightarrow M_{n \times n}(\mathbb{C})$$ $$A \mapsto (I+A^{-1})^{-1}+(I+A)^{-1}.$$

It follows from the chain rule and the well-known formula for the derivative of the inverse mapping that

$$D_Xf(\cdot) =-(I+X^{-1})^{-1}(-X^{-1}\cdot X^{-1})(I+X^{-1})^{-1}-(I+X)^{-1}\cdot (I+X)^{-1}.$$

$$=(X+I)^{-1}\cdot(X+I)^{-1}-(I+X)^{-1}\cdot (I+X)^{-1}=0.$$

$\mathcal{N}$ is connected. A proof of this is outlined below.

It suffices to prove that each matrix has a path to $I$. Therefore, take a matrix $A \in \mathcal{N}$. Due to density of matrices with different eigenvalues and openness of $\mathcal{N}$, there exists a matrix $K=PDP^{-1}$ near $A$ that can be joined to $A$ by a segment. Now, join $PDP^{-1}$ to $I$ by sending the diagonal entries to $1$ through a path which does not go through $0$ or $-1$ (using path-connectedness of $\mathbb{C}\backslash \{0,-1\})$.

It then suffices to compute $f$ in any given point. We compute $f(I)$:

$$f(I)= (2I)^{-1}+(2I)^{-1}=\frac{1}{2}I+\frac{1}{2}I=I.$$

Hence, $f$ is constant and equal to $I$. $\blacksquare$

(Note that since we've proved the proposition for matrices with complex entries, this holds also for real entries as well. It is also worth noting that I don't see how it would be possible to generalize this technique to an arbitrary field, since we heavily use differentiable properties and the topology of $\mathbb{C}$.)

*Thanks to Aaron for helpful comments regarding this answer!

Solution 3:

Write $B=(I_{n}+A^{-1})^{-1}+(I_{n}+A)^{-1}$. Can you compute the product $(I_n+A^{-1})B(I_n+A)$? If you can show that $$(I_n+A^{-1})B(I_n+A)=(I_n+A^{-1})(I_n+A),$$ then multiplying both sides by $(I_n+A^{-1})^{-1}$ on the left and $(I_n+A)^{-1}$ on the right, you get $B=I_n$.

Solution 4:

Actually there is an easier way to do that, just using basic matrix calculus - the argument by Aloizio Macedo however seems to be really elegant though. Let's suppose we already guessed that the following solution holds $$ (I_{n}+A^{-1})^{-1}+(I_{n}+A)^{-1}=I_n $$ for all suitable $A$. It is reasonable to assume this because, as usual, we just make some educated guessing with $A=I_n$, and $A$ some diagonal matrices in general.

Next step is we just verify this by doing the following steps \begin{align} & & (I_{n}+A^{-1})^{-1}+(I_{n}+A)^{-1}&=I_n \\ &\iff & (I_{n}+A^{-1})^{-1}(I_{n}+A)+I_n&=(I_{n}+A) \\ &\iff & (I_{n}+A^{-1})^{-1}(I_{n}+A)&=A \\ &\iff & I_n(I_{n}+A)&=(I_{n}+A^{-1})A \\ &\iff & (I_{n}+A)&=(A+I_n)\\ &\iff & A&=A \end{align} which is a true statement. Therefore $$ (I_{n}+A^{-1})^{-1}+(I_{n}+A)^{-1}=I_n $$ holds indeed for all suitable $A$.