Does the Cayley–Hamilton theorem work in the opposite direction?

Question

The Cayley–Hamilton theorem states that every square matrix satisfies its own characteristic equation.

But does it work in the opposite direction?

If for example for a certain matrix $A$ we know that

$ A^2-6A+9I=0, $

does that mean that the characteristic equation of $A$ is

$ \lambda^2-6\lambda+9=0 $ ?

Answer 1

Even without counterexamples it is obvious that your statement can't be true because if $A$ is a root of the polynomial $p(x)$ then it must be the root of $p(x)q(x)$ for any polynomial $q$. So that way we would get the matrix $A$ has infinitely many characteristic polynomials.

Answer 2

No, the $n\times n$ matrix $A=3I$ satisfies the given equation but it has a different characteristic polynomial for $n\not=2$.

Answer 3

No. For example, $I-1=0$, but the characteristic polynomial of $I$ is $(x-1)^n$.

Answer 4

For any $n \times n$ matrix $A$, the $(2n) \times (2n)$ matrix $\pmatrix{A & 0\cr 0 & A\cr}$ satisfies the characteristic polynomial of $A$, but its own characteristic polynomial is the square of that of $A$.

Answer 5

No. In general if a matrix is a root of a polynomial, that polynomial is a multiple of the minimal polynomial of that matrix.

Even if you only consider n-degree polynomials for an n×n matrix, this still can fail.

For example, if A=0, then the characteristic polynomial is $x^n$, but A is the root of any polynomial $xg(x)$.

However, all is not lost. If your minimal polynomial equals your characteristic polynomial (as with companion matrices or in the case you have no repeated roots), then the characteristic polynomial (= minimal polynomial) will be the unique monic n-degree polynomial with root A.