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Can an $n \times n$ matrix satisfy an $n$ degree polynomial equation other than its characteristic polynomial equation?

I was curious if the characteristic polynomial equation is the only $n$ degree equation that can be satisfied by a matrix. I have tried by trial and error to make up an equation for $2\times 2$ matrix but always end up with the characteristic polynomial.

Batominovski
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Aspirant
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    For a square matrix $X$ over a field $\mathbb{K}$ and for a polynomial $p(t)\in\mathbb{K}[t]$, we have $p(X)=0$ if and only if $\mu(t)$ divides $p(t)$, where $\mu(t)$ is the minimal polynomial of $X$. Hence, for your question, a matrix can be annihilated by more than one monic polynomial of degree $n$ if and only if its minimal polynomial has degree less than $n$, which is equivalent to saying that there exists an eigenvalue $\lambda$ of $X$ in the algebraic closure of $\mathbb{K}$ with geometric multiplicity greater than $1$. – Batominovski Jul 06 '20 at 18:42
  • To add to Batominovski's comment, here is a list of conditions equivalent to the condition $p(t) = \mu(t)$. See also this post. – Ben Grossmann Jul 06 '20 at 18:58
  • @Batominovski So for a 2 X 2 matrix (except the identity matrix) having eigenvalues 1,1 is it necessary for the matrix to satisfy a two degree monic polynomial (X-1)(X-K) for some real K (K is not equal to 1) (the matrix clearly cannot satisfy a monomial) – Aspirant Jul 07 '20 at 04:20
  • @KushalGupta I don't understand the question in your comment above, but the only $2$-by-$2$ matrix $X$ with only one eigenvalue $1$ and with minimal polynomial $\mu(t)=t-1$ is the identity matrix. – Batominovski Jul 07 '20 at 10:22
  • @Batominovski let me reframe: For a 2×2 matrix having eigenvalues 1,1 will the matrix satisfy a two degree monic polynomial other than characteristic polynomial?If yes,shouldn't one of the roots of the polynomial be the eigenvalue (1)? – Aspirant Jul 07 '20 at 15:52

3 Answers3

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The $2\times 2$ identity matrix satisfies $X(X-1)$.

ancient mathematician
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If the eigenvalues are distinct, then the characteristic polynomial is the only monic $n$th degree polynomial the matrix satisfies, since the eigenvalues are the roots, and so determine an $n$th degree polynomial up to a constant factor. If the minimal polynomial is the characteristic polynomial, we can make a similar statement. If not, then it's not true as ancientmathematician's answer shows.

saulspatz
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The question (supposing you implicitly don't want to talk about scalar multiples of the characteristic equation) is whether there exist, in the degree $n$ of the characteristic polynomial, any polynomial multiples of the minimal polynomial (i.e., annihilating polynomials) that are not (necessarily scalar) multiples of the characteristic polynomial. It is easy to see that the reply is: such multiples exists if and only if the minimal polynomial is of degree strictly less than $n$ (just multiply it by a factor of the necessary degree that is not a scalar multiple of the exact quotient of the characteristic polynomial by the minimal polynomial; this is clearly possible once that quotient is not the constant polynomial$~1$. Examples are easily found; particularly easy ones involve scalar multiples of the identity matrix, whose minimal polynomial is of degree$~1$.

Marc van Leeuwen
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