Samme013 said:
I am not sure if i can use that fact and i can't really ask but it is kind of trivial if i can use it.Is it even possible to show it without using that?
It might be. You might be able to do something clever with determinants. Let me marshall the facts such as we know them:
\begin{align*}
&\text{Idempotent matrices have determinant equal to } 0 \text{ or } 1. \text{ Why?} \\
A^2&=A \\
(\det(A))^2&=\det(A) \\
(\det(A))^2-\det(A)&=0 \\
\det(A)(\det(A)-1)&=0 \\
\det(A)&\in\{0,1\} \\
\therefore \; \det(I-A)&\in\{0,1\}. \\
&\text{Idempotent matrices are either the identity matrix, or are singular.} \\
&\text{Why? Assume } A \text{ is non-singular. Then } \\
A^2&=A \\
A^{-1}AA&=A^{-1}A \\
IA&=I \\
A&=I.
\end{align*}
Now we could take two cases: 1. $A=I$. 2. $\det(A)=0$.
For the first case, $I+A=I+I=2I$, and the determinant is $2^n\not=0$, where $A$ is $n\times n$.
For the second case, we have $\det(A)=0$. What if you formed the product
$$(I+A)(I+A)=I^2+IA+AI+A^2=I+2A+A=I+3A.$$
Not sure where that leads you. You could also form
$$(I-A)(I+A)=I^2+IA-AI-A^2=I+A-A-A=I-A.$$
Just seeing here:
$$(I-A)(I-A)=I^2-IA-AI+A^2=I-A-A+A=I-A.$$
Evidently, $(I-A)(I-A)=(I-A)(I+A)$, but I don't see where you can go further. There's no guarantee that $I-A$ is invertible (indeed, unless $I-A=I$, it won't be!), so you can't cancel anything, and the determinants won't give you anything here, either.
I think that using the formula for the inverse the problem asks you to show is just fine. If you simply compute $(I+A)(I-\tfrac12 A)=I$, you have just shown that $I+A$ is invertible. The problem asks you to show those two things. It does not tell you in what order to do them.
