How to find ${\rm rank}(2I_n-A)$ where A is a square matrix of size $n$ and $A^3 - 6A^2 + 12A = 0_n$?

Question

I think the rank has to be $n$ since anything else would be impossible to prove with so little information about the matrix.

$$\det(2I_n-A) = -P_A(2) = -\det(A - 2I_n) \ .$$

So, if I can show the characteristic polynomial is non-zero at $2$, it would prove the rank is $n$. I have no idea how to do that though. Maybe the characteristic polynomial can be deduced from

$$A^3 - 6A^2 + 12A = 0$$

somehow? Not sure how, since the Hamilton-Cayley theorem doesn't work in the opposite direction.

If I factorize the polynomial I get

$$(A^2 - 6A + 12I_n)A = 0_n \ .$$

So, at least one of the factors has to have the determinant $0$, but I want to prove that $A$'s determinant isn't $0$, so I'm not sure if factorizing helps here.

Answer 1

$P_A(x)=x^3-6x^2+12x$ is an annihilating polynomial of $A$.

Hence $\textrm{Spec}(A) \subset \{\lambda: P_A(\lambda)=0\}$

Since $P_A(2) =8\neq 0$ , so $2$ is not an eigenvalue of $A$.

Hence $\det(2I_n-A) \neq 0$ implies $2I_n-A$ is invertible, hence $\textrm{rank}(2I_n-A) =n$

Answer 2

From $p(A)=0$ you don't get that $p$ is the characteristic polynomial for $A$. What you do get is that $p$ is a multiple of the minimal polynomial for $A$. And any eigenvalue is a root of the minimal polynomial.

As $2$ is not a root of $p$ (roots are $0$ and two conjugate non-real complex numbers), it is not an eigenvalue of $A$ and hence $2I_n-A$ is invertible and thus has full rank.

Answer 3

No need here to reason on minimal or characteristic polynomial, nor eigenvalues.

Simply let $$B:=2I_n-A,$$ and notice that $$B^3=-A^3+6A^2-12A+8I_n=8I_n,$$ so $$\operatorname{rank}(B)=n,$$ since $B$ is invertible ($B^{-1}=\frac18B^2$).