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

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

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

• Yes there's an “opposite direction” that works, namely that the minimal polynomial (which divides the characteristic polynomial and often equals it) of $A$ divides (rather than necessarily equals) $\lambda^2 - 6\lambda + 9 = 0$. See en.wikipedia.org/wiki/Minimal_polynomial_(linear_algebra) – ShreevatsaR Feb 5 at 21:41
• The answers would be a lot more instructive if the question had a condition that the degree of the polynomial is the same as the size of the matrix. Too late now, I'm afraid. – JiK Feb 6 at 0:23

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.

• True, thank you – Ido Feb 5 at 20:40
• Are all counterexamples of this form? – PyRulez Feb 6 at 0:26
• @PyRulez, the set of polynomials $p$ such that $p(A)=0$ is $\{mq$: q is a polynomial$\}$ when $m$ is the minimal polynomial of $A$. So all such polynomials are multiples of the minimal polynomial. – Mark Feb 6 at 0:39

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

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

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 $$A$$.

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.