Elegant proof for a statement

Question

Given the matrix $$A=\begin{pmatrix} 0 & 1 & 1\\ 1 & 0 & 1\\ 1 & 1 & 0 \end{pmatrix}$$ Is there an elegant way to prove that $$(A-I_{3})^{-1}-A^{-1}=(A-I_{3})^{-1}A^{-1} ?$$ Note: It's easy to just calculate it, but I suspect it can be done by a nice approach.

Answer 1

$$(A-I)^{-1}(A-I) = I \Rightarrow\\ (A-I)^{-1}A - (A-I)^{-1} = I \Rightarrow\\ (A-I)^{-1}A - (A-I)^{-1} = A^{-1}A \Rightarrow\\ \left[(A-I)^{-1}-A^{-1}\right]A = (A-I)^{-1} \Rightarrow \\ \left[(A-I)^{-1}-A^{-1}\right]AA^{-1} = (A-I)^{-1}A^{-1} \Rightarrow \\ (A-I)^{-1}-A^{-1} = (A-I)^{-1}A^{-1}.$$

Notice that this does not depend upon the the dimension on $A$, nor on its particular form. It is only asked that both $A$ and $A-I$ are invertible.

Answer 2

Multiply on both sides from the left by $A(A-I)$. Then multiply from the right by $A(A-I)$, and you get something you can confirm without calculating any inverses.

Answer 3

\begin{align}(A-I_{3})^{-1}-A^{-1}=(A-I_{3})^{-1}A^{-1}&\iff(A-I_3)(A-I_3)^{-1}A-(A-I_3)A^{-1}A=I_3\\&\iff A-(A-I_3)=I_3.\end{align}

Answer 4

$A+I$ has rank $1$ and eigenvalues $0,0,3$. It follows that $A$ has eigenvalues $-1,-1,2$ and the characteristic polynomial of $A$ is given by $(x+1)^2(x-2)$. In particular neither $0$ or $1$ belong to the spectrum of $A$ and $$ \frac{1}{A-I}-\frac{1}{A} = \frac{1}{A(A-I)} $$ holds as a consequence of $\frac{1}{x-1}-\frac{1}{x}=\frac{1}{x(x-1)}$ for any $x\not\in\{0,1\}$.

Answer 5

It means that $\;(A-I)^{-1}-A^{-1}=\bigl(A(A-I)\bigr)^{-1}$, so compute the product $$A(A-I)\bigl((A-I)^{-1}-A^{-1}\bigr)= A(A-I)(A-I)^{-1}-A(A-I)A^{-1}=A-A(A-I)A^{-1}.$$ Now note $A$ and $A-I$ commute, so $$A-A(A-I)A^{-1}=A-AA^{-1}(A-I)=A-(A-I)=I.$$ Similar computation for $\;\bigl((A-I)^{-1}-A^{-1}\bigr)A(A-I)$.