Determine what condition the $n\times n$ matrix $A$ must fulfill so that the matrix $I_n - A$ has an inverse and this equals $I_n + A$.

Question

Determine what condition the $n\times n$ matrix $A$ must fulfill so that the matrix $I_n - A$ has an inverse and this equals $I_n + A$.

I tried to apply it to a $2\times 2$ matrix obtaining the following result:

$A = \begin{pmatrix} a & b\\ c & d \end{pmatrix}; $

$(I_n-A)^{-1} = \frac{1}{(1-a)(1-d)-bc} * \begin{pmatrix} 1-d & b\\ c & 1-a \end{pmatrix} = \begin{pmatrix} 1+a & b\\ c & 1+d \end{pmatrix} $

Answer 1

$(I - A)^{-1} = I + A \Longleftrightarrow A^2 = 0; \tag 1$

for if

$A^2 = 0, \tag 2$

then

$(I - A)(I + A) = I + A - A - A^2 = I - A^2 = I; \tag 3$

likewise if

$(I - A)(I + A) = I, \tag 4$

then

$I - A^2 = (I - A)(I + A) = I, \tag 5$

then

$I = A^2 + I, \tag 6$

or

$A^2 = 0. \tag 7$

Answer 2

If the inverse should be $I_n+A$ you need that $$ (I_n-A)(I_n+A)=I_n $$ What do you get?

Answer 3

Hint:$$(I-A)(I+A)=I\color{blue}{-A^2}$$

Answer 4

More generally, if the matrix $A$ is nilpotent, i.e. if $A^n=0$ for some $n>0$, then $I-A$ is invertible, and its inverse is $$(I-A)^{-1}=I+A+\dots+A^{n-1}.$$