The Minimal Polynomial and Its Existence

From an Annihilating Polynomial to the Minimal One

A polynomial can encode a relation among the powers of a linear operator. For example, if \(T^2-5T+6\operatorname{id}_V\) is the zero operator, then every higher power of \(T\) can be reduced using that relation. The minimal polynomial captures the simplest such relation: it is the monic polynomial of least degree that evaluates to the zero operator. This tutorial establishes that such a polynomial exists and is unique for every linear operator on a finite-dimensional space.

Let \(V\) be a nonzero finite-dimensional vector space over a field \(\mathbb{F}\), where \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{C}\), and let \(T:V\to V\) be linear. A polynomial \(f\in\mathbb{F}[x]\) has the form \(f(x)=a_0+a_1x+\cdots+a_kx^k\). Its evaluation at \(T\) is the operator \(f(T)=a_0\operatorname{id}_V+a_1T+\cdots+a_kT^k\), where \(T^0=\operatorname{id}_V\). We call \(f\) an annihilating polynomial for \(T\) if \(f(T)\) is the zero operator.

The condition is about the operator on every vector in \(V\), not about the value of a polynomial at one vector. In particular, \(f(T)=0\) means \(f(T)\mathbf{v}=\mathbf{0}\) for every \(\mathbf{v}\in V\). We use the zero operator symbolically as \(0\) in polynomial identities involving operators.

Polynomial evaluation respects addition and multiplication: \((f+g)(T)=f(T)+g(T)\) and \((fg)(T)=f(T)g(T)\). To see the multiplication rule, expand \(f\) and \(g\) into powers of \(x\). Each product of terms \(a_iT^i\) and \(b_jT^j\) is \(a_ib_jT^{i+j}\), since \(T^iT^j=T^{i+j}\). These are exactly the terms obtained by evaluating the product polynomial. We will use these identities when applying polynomial division.

Why an Annihilating Polynomial Exists

Choose a basis of \(V\), let \(n=\dim V\), and let \(\mathbf{A}\) be the matrix of \(T\) in that basis. The characteristic polynomial convention here is \(p_{\mathbf{A}}(x)=\det(\mathbf{A}-x\mathbf{I}_n)\), whose leading coefficient is \((-1)^n\). Thus \(q(x)=(-1)^n p_{\mathbf{A}}(x)\) is monic and has degree \(n\). The Cayley–Hamilton theorem states that \(p_{\mathbf{A}}(\mathbf{A})=\mathbf{O}\), so also \(q(\mathbf{A})=\mathbf{O}\). Since evaluating a polynomial at the matrix of \(T\) gives the matrix of that polynomial in \(T\), \(q(T)\) is the zero operator. Cayley–Hamilton therefore supplies at least one monic annihilating polynomial.

Proof. By Cayley–Hamilton, the set of nonzero annihilating polynomials is nonempty. Their degrees are nonnegative integers, so this set of degrees has a least element \(d\). Choose an annihilating polynomial \(f\) of degree \(d\). Its leading coefficient \(a\) is nonzero, and \(a^{-1}f\) is also an annihilating polynomial, now monic and still of degree \(d\). This proves existence.

For uniqueness, suppose \(m_1\) and \(m_2\) are monic annihilating polynomials of least degree. Divide \(m_1\) by \(m_2\) using polynomial division over the field \(\mathbb{F}\): \(m_1=qm_2+r\), where either \(r=0\) or \(\deg r<\deg m_2\). Evaluating at \(T\) and using the addition and multiplication rules gives \(0=m_1(T)=q(T)m_2(T)+r(T)=r(T)\). If \(r\ne0\), then dividing \(r\) by its nonzero leading coefficient produces a monic annihilating polynomial of degree less than \(\deg m_2\), contradicting the least-degree property. Hence \(r=0\). Therefore \(m_1=qm_2\). Since both polynomials are monic and have the same degree, \(q\) is the constant polynomial \(1\), and \(m_1=m_2\). \(\square\)

The finite-dimensional hypothesis is used when we invoke Cayley–Hamilton. The field hypothesis is used for polynomial division and for scaling a polynomial by its leading coefficient. If \(V\) is the zero vector space, every operator on \(V\) is the zero operator, including \(\operatorname{id}_V\). In that special case the unique monic polynomial of least degree that annihilates \(T\) is the constant polynomial \(1\).

Computing and Checking Examples

The Operator and Its Matrix

The definition belongs to the linear operator, even though calculations are often done with a matrix. A change of basis changes the matrix, but not which polynomials annihilate the operator.

Proof. The two matrices are related by a change of basis: \(\mathbf{B}=\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\) for an invertible matrix \(\mathbf{P}\). For every integer \(k\geq 0\), \(\mathbf{B}^k=\mathbf{P}^{-1}\mathbf{A}^k\mathbf{P}\). For \(k=0\), both sides equal \(\mathbf{I}\); for positive \(k\), the identity follows by multiplying the expression for \(\mathbf{B}\) repeatedly and using \(\mathbf{P}\mathbf{P}^{-1}=\mathbf{I}\). If \(f(x)=\sum_{k=0}^s a_kx^k\), it follows that \(f(\mathbf{B})=\sum_{k=0}^s a_k\mathbf{B}^k=\mathbf{P}^{-1}(\sum_{k=0}^s a_k\mathbf{A}^k)\mathbf{P}=\mathbf{P}^{-1}f(\mathbf{A})\mathbf{P}\). This is zero exactly when \(f(\mathbf{A})\) is zero, because \(\mathbf{P}\) is invertible. The annihilating polynomials, and therefore the unique monic one of least degree, are independent of the chosen basis. \(\square\)

Common Pitfall: Cayley–Hamilton Does Not Give Minimality

Cayley–Hamilton guarantees that the characteristic polynomial annihilates a matrix; it does not say that this polynomial has the least possible degree. For example, let \(\mathbf{A}=\begin{bmatrix}3&0\\0&3\end{bmatrix}\). Its characteristic polynomial under the course convention is \(p_{\mathbf{A}}(x)=\det(\mathbf{A}-x\mathbf{I}_2)=(3-x)^2\). A mistaken conclusion is that the minimal polynomial must have degree \(2\), since the matrix is \(2\times2\). But

Therefore \(x-3\) is a monic degree-one annihilating polynomial. The monic degree-zero polynomial \(1\) evaluates to \(\mathbf{I}_2\ne\mathbf{O}\), so \(m_T(x)=x-3\). The correction is to use Cayley–Hamilton for existence, then check whether a lower-degree monic annihilating polynomial is possible; annihilation alone does not establish minimality.

The distinction between “annihilating” and “minimal” is central. An operator can have many annihilating polynomials, including multiples of any one that annihilates it. The minimal polynomial selects the unique monic one with least degree. The existence proof relies on finite-dimensionality through Cayley–Hamilton; the uniqueness proof relies on polynomial division and the least-degree condition. These arguments establish the definition without requiring any particular matrix form.