I just can't wrap my head around this proof. How to do it? Here is what I tried:
$$A^3 = A$$ $$A^3 - A = 0$$ $$A(A-I)(A+I) = 0$$ Matrix $A$ satisfies the polynomial $p(x)=x(x-1)(x+1)$. Zeroes of $p$ are $-1, 0 ,1$. Minimal polynomial of the matrix $A$ has at least one root, which can only take value of roots of $p$. Maximum multiplicity of that root in the minimal polynomial is one.
One can then take a look at the eigenspaces that correspond to the specific eigenvalues. What then...? Is that the path that can lead to a solution? Is there a better solution? I don't really get it... Please, be kind and explain this if you can spare the time.
I found this question, but there seems to be a leap of logic in the question, and the accepted answer.
$$A^3 = A$$ doesn't imply $$A^2 = I$$
For instance, take matrix $$ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} $$
Also, if we take a linear operator $L:V \to V$, it is obvious that $Ker(L) \cap Im(L) = \{0\}$ ? Do I also need to prove that sum of kernel and image is direct sum?
I assume the space is finite dimensional.
Let $y\in \operatorname{Ker}(A-I)\cap \operatorname{Im}(A-I)$, $y=A(x)-x, (A-I)(y)=0$ implies that $(A-I)(A(x)-x)=A^2(x)-2A(x)+x=0$, we deduce that $A^3(x)-2A^2(x)+A(x)=A(x)-2A^2(x)+A(x)=0$. This implies that $A^2(x)=A(x)$,
we replace $A(x)=A^2(x)$ in $A^2(x)-2A(x)+x=0$, we obtain $A(x)-2A(x)+x=x-A(x)=-y=0$
Since $\dim(\operatorname{Ker}(A-I)+\dim(\operatorname{Im}(A-I))=\dim(V)$ the result follows.