Matrix Polynomials and the Minimal Polynomial

Definition Write $\mathbb{F}[t]$ for the polynomials in one variable $t$ with coefficients in $\mathbb{F}$. Powers of $T$ are defined by $T^0 = I$ and $T^{k+1} = T \circ T^k$. For $f(t) = a_0 + a_1 t + \cdots + a_d t^d$ in $\mathbb{F}[t]$, the polynomial $f$ evaluated at $T$ is the operator

The same formula with $I = I_n$ defines $f(A)$ for a square matrix $A \in \mathbb{F}^{n \times n}$. For all $f, g \in \mathbb{F}[t]$ and $c \in \mathbb{F}$, evaluation obeys

The product rule holds because $T^j T^k = T^{j+k}$, so expanding $f(T)\,g(T)$ and collecting the terms with the same total exponent is the same bookkeeping as multiplying the scalar polynomials. Since $fg = gf$, any two polynomials in the same operator commute, $f(T)\,g(T) = g(T)\,f(T)$. If $A$ is the matrix of $T$ in a basis $\mathcal{B}$ of $V$, then the matrix of $f(T)$ in that basis is $f(A)$. For an invertible $S \in \mathbb{F}^{n \times n}$ the powers satisfy $(S^{-1}AS)^k = S^{-1}A^k S$, hence $f(S^{-1}AS) = S^{-1} f(A)\, S$.

Definition A polynomial $f \in \mathbb{F}[t]$ is an annihilating polynomial of $T$, or annihilates $T$, if $f \ne 0$ and $f(T)$ is the zero operator, that is, $f(T)v = 0$ for every $v \in V$. Every operator on a finite-dimensional space has one. The space $\operatorname{Hom}(V, V)$ has dimension $n^2$, so the $n^2 + 1$ operators

are linearly dependent. A nontrivial relation $c_0 I + c_1 T + \cdots + c_{n^2} T^{n^2} = 0$ gives the nonzero polynomial $f(t) = c_0 + c_1 t + \cdots + c_{n^2} t^{n^2}$ with $f(T) = 0$. Because $V \ne \{0\}$, a nonzero constant $c$ gives $cI \ne 0$, so every annihilating polynomial has degree at least $1$. If $f$ annihilates $T$, so does $gf$ for every nonzero $g \in \mathbb{F}[t]$, since $(gf)(T) = g(T)\,f(T) = 0$.

Definition The division theorem says that for $f, m \in \mathbb{F}[t]$ with $m \ne 0$ there are unique $q, r \in \mathbb{F}[t]$ with

and either $r = 0$ or $\deg r < \deg m$. For existence, as long as the current remainder has degree at least $\deg m$, subtract the multiple $b\,t^k m$ that cancels its leading term, where $k$ is the difference of the degrees and $b$ the ratio of the leading coefficients. Each step lowers the degree, so the process stops. For uniqueness, suppose $qm + r = q'm + r'$ with both remainders as required. Then $(q - q')m = r' - r$. If $q \ne q'$, the left side has degree at least $\deg m$ while the right side is zero or has smaller degree, which is impossible. So $q = q'$ and then $r = r'$. The division is carried out with $t$ as a scalar variable, and only afterwards is the identity evaluated at $T$, which the evaluation rules allow, giving $f(T) = q(T)\,m(T) + r(T)$. Dividing by $m(t) = t - c$ leaves the constant remainder $f(c)$, so $f(c) = 0$ exactly when $t - c$ divides $f$.

Definition The minimal polynomial $\mu_T$ of $T$ is the monic annihilating polynomial of least degree, where monic means that the leading coefficient is $1$. It exists, because annihilating polynomials exist and dividing one of least degree by its leading coefficient keeps it annihilating. It is unique. If two different monic annihilators had the least degree $d$, their difference would be nonzero, of degree less than $d$, and still annihilating, which contradicts the choice of $d$. The divisibility theorem says that for every $f \in \mathbb{F}[t]$

To prove it, divide $f = q\,\mu_T + r$ with $r = 0$ or $\deg r < \deg \mu_T$. Evaluating at $T$ gives $0 = q(T)\,\mu_T(T) + r(T) = r(T)$. A nonzero $r$ would be an annihilating polynomial of smaller degree than $\mu_T$, so $r = 0$. Conversely, $f = q\,\mu_T$ gives $f(T) = q(T)\,\mu_T(T) = 0$. For a matrix $A \in \mathbb{F}^{n \times n}$ the minimal polynomial $\mu_A$ is defined the same way. Similar matrices have the same minimal polynomial, because $f(S^{-1}AS) = S^{-1} f(A) S$ is zero exactly when $f(A)$ is. Since the matrix of $f(T)$ is $f(A)$, the operator $T$ and every matrix $A$ of $T$ have the same annihilating polynomials, so $\mu_T = \mu_A$.

Definition The roots theorem says that for every $\lambda \in \mathbb{F}$, $\mu_T(\lambda) = 0$ if and only if $\lambda$ is an eigenvalue of $T$, that is, $Tv = \lambda v$ for some nonzero $v \in V$. If $Tv = \lambda v$ with $v \ne 0$, then $T^k v = \lambda^k v$ for every $k$, and therefore every $f \in \mathbb{F}[t]$ satisfies

With $f = \mu_T$ the left side is $0$, and $v \ne 0$ forces $\mu_T(\lambda) = 0$. Conversely, if $\mu_T(\lambda) = 0$, then $\mu_T(t) = (t - \lambda)\,q(t)$ with $q$ monic of degree $\deg \mu_T - 1$, and

If $T - \lambda I$ were invertible, multiplying by its inverse would give $q(T) = 0$, contradicting the minimality of $\mu_T$. So $T - \lambda I$ is not invertible, and on the finite-dimensional space $V$ this means its kernel contains a nonzero vector, an eigenvector for $\lambda$. Consequently the roots of $\mu_T$ in $\mathbb{F}$ are exactly the roots of $\chi_T$ in $\mathbb{F}$, although their multiplicities may differ. Only roots in the stated field count.