1. The problem is to analyze the matrix equation $$A^3 - (A^2 - A) = 0$$ for a matrix $A \in M_n(\mathbb{R})$ with rank $\mathrm{rk}(A) = M$ and to deduce properties about $A$ and its spectrum. 2. Given the equation, rewrite it as $$A^3 = A^2$$. 3. For an eigenvalue $\lambda \in \mathrm{Sp}(A)$ with eigenvector $X \neq 0$, we have $$A X = \lambda X$$. 4. Applying powers of $A$ to $X$ gives $$A^2 X = \lambda^2 X$$ and $$A^3 X = \lambda^3 X$$. 5. Substitute into the equation: $$(A^3 - A^2) X = 0 \implies (\lambda^3 - \lambda^2) X = 0$$. 6. Factor the eigenvalue equation: $$\lambda^2 (\lambda - 1) = 0$$ which implies $$\lambda = 0 \text{ or } \lambda = 1$$. 7. Therefore, the spectrum satisfies $$\mathrm{Sp}(A) \subset \{0,1\}$$. 8. Since $\mathrm{rk}(A) = \sum_{\lambda \in \mathrm{Sp}(A)} \lambda = n$ and $\mathrm{Sp}(A) \subset \{0,1\}$, if $0 \in \mathrm{Sp}(A)$ then $$\mathrm{rk}(A) \leq n-1$$ which contradicts the rank condition. 9. Hence, $$0 \notin \mathrm{Sp}(A) \implies A \text{ is invertible}$$. 10. Using $$A^3 = A^2$$ multiply both sides by $$(A^{-1})^2$$ to get $$A = I_n$$. 11. If $A$ is diagonalizable, then $$A = I_n$$ and $$\mathrm{Sp}(A) = \{1\}$$. Final answer: $$A = I_n$$.