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## Rank Relations

Suppose we have a matrix $A \in \mathcal{M}_n(\mathbb{R})$ such that $A^3=A$. Prove that $rank(A) +rank (A- I_n)+rank (A + I_n)=2n$.

As a generalization, we can state the following:
Given a matrix $A$ with $p(A)=0$, where $p$ is a real polynomial with distinct roots, we have $\sum_{j=1}^k rank(A- x_j I) =(k-1)\cdot n$, and $x_i$ are the distinct roots of $p$.

Solution:
Since $p(A)=0$ and $p$ has distinct real roots it follows that the minimal polynomial of $A$ has distinct linear factors, implying that $A$ is diagonalizable. Therefore, there exists a non-singular matrix $P$ such that $A=PDP^{-1}$, where $D$ is a diagonal matrix having the eigenvalues of $A$ as entries on the diagonal.
Since $p(A)=0$ we can see that any root of $p$ is an eigenvalue for $A$. Therefore, on the diagonal of $D$ we have only roots of $p$.
Furthermore, we know that if $B$ is a matrix and $C$ is a nonsingular matrix, then $rank(B)=rank(CB)=rank(BC)$, so we can say that $rank(A-x_j I)=rank(PDP^{-1}-x_j I)=rank(D-x_j I)=$ number of diagonal entries not equal to $x_j$.
Adding up these relations we get the desired result.