$A$ is a $12\times12$ matrix such that $A^{10 }=I$, and $\operatorname{rank}(A-I)=5$. How can we show $\operatorname{rank}(A^2 +A+I)\leq 7$?

Question

If $A$ is a $12\times12$ matrix such that $A^{10 }=I$, and $\operatorname{rank}(A-I)=5$, how can we show that $\operatorname{rank}(A^2 +A+I)\leq 7$?

Answer 1

We can't because it isn't. The eigenvalues of $A$ are $10$'th roots of $1$. The roots of $\lambda^2+\lambda+1$ are primitive cube roots of $1$. So $A^2+A+I$ is nonsingular and has rank $12$.