$A_{n\times n}$ be a complex matrix with $A^2 = -I$ Let $A_{n\times n}$  be a complex matrix with $A^2 = -I$.
I want to prove that $A^3 = A^{-1}$, and disprove that $I - A = A^{-1}$.
I do not know if the matrix is invertible, else, this will simplify the rest of the work. Suppose that its not invertible, and suppose that i cant deduce from $A^2 + I = 0$ that $A$ is in polynomial vector space, particularly for $x^2 + 1$ dimension. How can I prove the first part?
For the second part:
$(1)\quad I - A = A^{-1} / \cdot A$
$(2)\quad A^{-1} - A^2 = I /$ substitute the given argument
$(3)\quad A^{-1} + I = I \quad \leftrightarrow \quad A^{-1} = 0$
Which is sort of "reductio ad absurdum" because the zero matrix is not invertible. Is this a correct proof?
Any advice and help would be highly appreciated!
 A: To summarize the comments, we have the following short answer.

*

*$A^2=-I$ implies that $A$ is invertible with $A=-A^{-1}$ and $A^4=I$. Hence we may multiply $A^4=I$ by $A^{-1}$ on both sides to obtain $A^3=A^{-1}=-A$.


*You can disprove $I-A=A^{-1}$ by choosing
$$
A=\begin{pmatrix} 0 & -1 \cr 1 & 0 \end{pmatrix}.
$$
Then $A^2=-I$, but $A^{-1}=-A\neq I-A$. For a $1\times 1$-matrix there is no real matrix $A$ with $A^2=-I$. Over the complex numbers, $A=(i)$ satisfies $A^2=-I$ and $I-A\neq A^{-1}$.
A: if $A^2=-I$, then $A^2+I=0$. By multyplying on both sides by $A$,we find $A^3+A=0\iff A^3=-A$.
Since $A^2=-I$ and $A$ is invertible because $\det(A^2)=\det(A)^2\neq0\implies \det(A)\neq 0$, we can multiply by the inverse (as it exists) on both sides and find $A=-A^{-1}\implies -A=A^{-1}$.
So finally:
$$A^3=-A=A^{-1}$$
Thus $A^3=A^{-1}$.
For the second part, since $-A=A^{-1}$ and $I\neq 0$, $I-A\neq A^{-1}$
A: If $A^2=-I$, then $A^4=I$, which directly implies that $A(A^3)=(A^3)A=I$; hence $A^3=A^{-1}$.
For the second part, $I-A \ne A^{-1}$ because
$$
      A(I-A)=(I-A)A= A-A^2 = A+I \ne I.
$$
