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 $ ?

  • 3
    $\begingroup$ 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) $\endgroup$ Feb 5 '19 at 21:41
  • 3
    $\begingroup$ 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. $\endgroup$
    – JiK
    Feb 6 '19 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.

  • $\begingroup$ True, thank you $\endgroup$
    – Ido
    Feb 5 '19 at 20:40
  • $\begingroup$ Are all counterexamples of this form? $\endgroup$
    – PyRulez
    Feb 6 '19 at 0:26
  • 2
    $\begingroup$ @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. $\endgroup$
    – Mark
    Feb 6 '19 at 0:39

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


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.


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


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