From Annihilating Polynomials to the Minimal Polynomial
Polynomial expressions in a linear operator let us record algebraic information about its powers. For example, an identity such as \(T^2-3T+2\operatorname{id}_V=0\) says that every higher power can be reduced to a combination of lower powers. The minimal polynomial captures the shortest monic polynomial identity of this kind. Two important questions are how it relates to all other polynomial identities for the same operator, and what its roots tell us about the operator’s eigenvalues.
Throughout, let \(V\) be a nonzero finite-dimensional vector space over \(\mathbb{C}\), and let \(T:V\to V\) be linear. For a polynomial \(p(x)=a_0+a_1x+\cdots+a_kx^k\) in \(\mathbb{C}[x]\), define its evaluation at \(T\) by \(p(T)=a_0\operatorname{id}_V+a_1T+\cdots+a_kT^k\), where \(T^0=\operatorname{id}_V\). We call \(p\) an annihilating polynomial for \(T\) if \(p(T)\) is the zero map. The minimal polynomial \(m_T\) is the monic annihilating polynomial of least degree. Its existence for a finite-dimensional space was established in the earlier material on the minimal polynomial.
In part (1), “divisible” means that \(p=m_Tq\) for some \(q\in\mathbb{C}[x]\). The zero polynomial is included: it is divisible by every nonzero polynomial. For part (2), an eigenvalue \(\lambda\) means that there is a vector \(\mathbf{v}\ne\mathbf{0}\) such that \(T(\mathbf{v})=\lambda\mathbf{v}\). The two statements have different jobs. Divisibility compares polynomial identities; the root statement identifies the eigenvalues from the minimal polynomial.
Why Every Annihilator Has the Minimal Polynomial as a Factor
The key is the polynomial division algorithm. Divide any annihilating polynomial \(p\) by \(m_T\), leaving a remainder whose degree is smaller than the degree of \(m_T\). If the remainder also annihilates \(T\), minimality forces it to be zero. Polynomial evaluation respects sums and products: powers of \(T\) satisfy \(T^iT^j=T^{i+j}\), so evaluating a product of polynomials gives the corresponding product of operators.
Proof. Suppose \(p(T)=0\). By polynomial division, there are \(q,r\in\mathbb{C}[x]\) such that \(p=q m_T+r\), where either \(r=0\) or \(\deg r<\deg m_T\). Evaluating at \(T\) gives \(0=p(T)=q(T)m_T(T)+r(T)=r(T)\), since \(m_T(T)=0\). If \(r\ne0\), its leading coefficient \(c\) is nonzero, and \((c^{-1}r)(T)=c^{-1}r(T)=0\). The polynomial \(c^{-1}r\) is monic and has degree smaller than \(m_T\), contradicting the defining minimality of \(m_T\). Therefore \(r=0\), so \(p=q m_T\), as required. \(\square\)
This proof uses minimality only after producing a nonzero annihilating polynomial of strictly smaller degree. It does not attempt to cancel a factor from a product of operators. That distinction matters: a nonzero operator need not have an inverse, so a matrix factor cannot generally be cancelled.
Why the Roots Are Exactly the Eigenvalues
First suppose \(\lambda\) is an eigenvalue, with eigenvector \(\mathbf{v}\ne\mathbf{0}\). For every nonnegative integer \(j\), repeated application of \(T\) gives \(T^j(\mathbf{v})=\lambda^j\mathbf{v}\). Thus evaluating any polynomial on this eigenvector amounts to evaluating the scalar polynomial at \(\lambda\). In particular, the identity \(m_T(T)=0\) forces \(m_T(\lambda)=0\).
For the converse, suppose \(m_T(\lambda)=0\). The factor theorem for polynomials gives \(m_T(x)=(x-\lambda)q(x)\), where \(q\ne0\) and \(\deg q=\deg m_T-1\). If \(\lambda\) were not an eigenvalue, the map \(T-\lambda\operatorname{id}_V\) would have zero kernel. Because \(V\) is finite-dimensional and this map goes from \(V\) to itself, the rank-nullity theorem implies that it is onto as well, hence invertible. But then its product with \(q(T)\) being zero would force \(q(T)=0\). That would contradict minimality.
Proof. If \(\lambda\) is an eigenvalue, choose \(\mathbf{v}\ne\mathbf{0}\) with \(T(\mathbf{v})=\lambda\mathbf{v}\). Induction on \(j\) gives \(T^j(\mathbf{v})=\lambda^j\mathbf{v}\): it holds for \(j=0\), and applying \(T\) to the equation for \(j\) gives the equation for \(j+1\). Writing \(m_T(x)=\sum_j a_jx^j\), we obtain \(0=m_T(T)(\mathbf{v})=\sum_j a_jT^j(\mathbf{v})=\sum_j a_j\lambda^j\mathbf{v}=m_T(\lambda)\mathbf{v}\). Since \(\mathbf{v}\ne\mathbf{0}\), \(m_T(\lambda)=0\).
Conversely, suppose \(m_T(\lambda)=0\). By the factor theorem, write \(m_T(x)=(x-\lambda)q(x)\), with \(q\ne0\) and \(\deg q<\deg m_T\). If \(\lambda\) is not an eigenvalue, the kernel of \(T-\lambda\operatorname{id}_V\) contains only \(\mathbf{0}\). Rank-nullity, applied to this endomorphism of a finite-dimensional space, then shows it is onto; a linear map that is both one-to-one and onto has an inverse. Evaluating the factorization gives \(0=m_T(T)=(T-\lambda\operatorname{id}_V)q(T)\). Multiplying by the inverse of \(T-\lambda\operatorname{id}_V\) yields \(q(T)=0\). Scaling \(q\) by its nonzero leading coefficient would give a monic annihilating polynomial of degree smaller than \(m_T\), a contradiction. Hence \(\lambda\) must be an eigenvalue. \(\square\)
Worked Examples
Worked Example: A Diagonal Matrix with Two Eigenvalues
Find the minimal polynomial of \(\mathbf{A}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&-1\end{bmatrix}\), and check an additional annihilating polynomial.
For a polynomial \(p\), evaluation on this diagonal matrix gives \(p(\mathbf{A})=\begin{bmatrix}p(2)&0&0\\0&p(2)&0\\0&0&p(-1)\end{bmatrix}\). Consequently, \(p(\mathbf{A})=\mathbf{O}\) exactly when \(p(2)=0\) and \(p(-1)=0\). The factor theorem says that an annihilating polynomial must contain both distinct factors \(x-2\) and \(x+1\). Their product is monic, so the candidate minimal polynomial is \(m_{\mathbf{A}}(x)=(x-2)(x+1)=x^2-x-2\).
Check the candidate directly. We have \(\mathbf{A}-2\mathbf{I}_3=\begin{bmatrix}0&0&0\\0&0&0\\0&0&-3\end{bmatrix}\) and \(\mathbf{A}+\mathbf{I}_3=\begin{bmatrix}3&0&0\\0&3&0\\0&0&0\end{bmatrix}\). Their product is \(\mathbf{O}\), so \(m_{\mathbf{A}}(\mathbf{A})=\mathbf{O}\). Every annihilating polynomial has both distinct factors, and therefore is divisible by their product. This verifies that no lower-degree monic annihilator exists.
For example, \(p(x)=(x-2)^2(x+1)(x+4)\) is annihilating and is divisible by \(m_{\mathbf{A}}\). Indeed, \(p(2)=0\) and \(p(-1)=0\), so \(p(\mathbf{A})=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=\mathbf{O}\). The roots of the minimal polynomial are \(2\) and \(-1\), the eigenvalues of this matrix.
Worked Example: A Nilpotent Matrix
For \(\mathbf{N}=\begin{bmatrix}0&1\\0&0\end{bmatrix}\), determine the minimal polynomial and its eigenvalues.
Multiplication gives \(\mathbf{N}^2=\begin{bmatrix}0&1\\0&0\end{bmatrix}\begin{bmatrix}0&1\\0&0\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}\). Thus \(x^2\) annihilates \(\mathbf{N}\). To rule out a smaller-degree monic annihilator, write any polynomial of degree at most one as \(p(x)=c_0+c_1x\). Then \(p(\mathbf{N})=c_0\mathbf{I}_2+c_1\mathbf{N}=\begin{bmatrix}c_0&c_1\\0&c_0\end{bmatrix}\). If this is \(\mathbf{O}\), its diagonal entries give \(c_0=0\), and its upper-right entry gives \(c_1=0\). There is no nonzero annihilator of degree less than two, so \(m_{\mathbf{N}}(x)=x^2\).
The only root of \(m_{\mathbf{N}}\) is \(0\), and it is an eigenvalue because \(\mathbf{N}\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\). To check that there are no other eigenvalues, suppose \(\mathbf{N}\mathbf{v}=\lambda\mathbf{v}\) for \(\mathbf{v}\ne\mathbf{0}\). Applying \(\mathbf{N}\) again gives \(\mathbf{0}=\mathbf{N}^2\mathbf{v}=\lambda^2\mathbf{v}\), so \(\lambda^2=0\) and \(\lambda=0\).
As another check of divisibility, \(p(x)=x^3(x-4)\) annihilates \(\mathbf{N}\), since \(p(\mathbf{N})=\mathbf{N}^3(\mathbf{N}-4\mathbf{I}_2)=\mathbf{O}\) and \(\mathbf{N}^2=\mathbf{O}\). It has the required factor \(x^2=m_{\mathbf{N}}(x)\).
Worked Example: A Repeated Root
Let \(\mathbf{J}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\). Find its minimal polynomial and verify that a repeated root corresponds to its eigenvalue.
Subtract the identity matrix: \(\mathbf{J}-\mathbf{I}_2=\begin{bmatrix}0&1\\0&0\end{bmatrix}\), and hence \((\mathbf{J}-\mathbf{I}_2)^2=\begin{bmatrix}0&1\\0&0\end{bmatrix}\begin{bmatrix}0&1\\0&0\end{bmatrix}=\mathbf{O}\). Thus \((x-1)^2\) annihilates \(\mathbf{J}\). If \(p(x)=c_0+c_1x\), then \(p(\mathbf{J})=\begin{bmatrix}c_0+c_1&c_1\\0&c_0+c_1\end{bmatrix}\). For this matrix to be \(\mathbf{O}\), its upper-right entry requires \(c_1=0\), after which a diagonal entry requires \(c_0=0\). Therefore no nonzero polynomial of degree below two annihilates \(\mathbf{J}\), and \(m_{\mathbf{J}}(x)=(x-1)^2\).
The root is \(1\), and \(\mathbf{J}\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix}\), so \(1\) is an eigenvalue. For an explicit divisibility check, take \(g(x)=(x-1)^2(x+2)\). Then \(g(\mathbf{J})=(\mathbf{J}-\mathbf{I}_2)^2(\mathbf{J}+2\mathbf{I}_2)=\mathbf{O}\begin{bmatrix}3&1\\0&3\end{bmatrix}=\mathbf{O}\), and \(g\) is divisible by \(m_{\mathbf{J}}\).
A Common Cancellation Error
A matrix product can be zero even though neither factor is zero. For a concrete example, let \(\mathbf{D}=\begin{bmatrix}1&0\\0&2\end{bmatrix}\). Then \((\mathbf{D}-\mathbf{I}_2)(\mathbf{D}-2\mathbf{I}_2)=\begin{bmatrix}0&0\\0&1\end{bmatrix}\begin{bmatrix}-1&0\\0&0\end{bmatrix}=\mathbf{O}\). A wrong calculation cancels the first factor and concludes that \(\mathbf{D}-2\mathbf{I}_2=\mathbf{O}\). The correction is that \(\mathbf{D}-\mathbf{I}_2\) is not invertible: it sends the nonzero vector \(\begin{bmatrix}1\\0\end{bmatrix}\) to \(\mathbf{0}\). Cancellation is justified only for an invertible factor. In the root proof, the factor \(T-\lambda\operatorname{id}_V\) is cancellable only after assuming that \(\lambda\) is not an eigenvalue and using finite-dimensionality to establish invertibility.
The divisibility result gives another safe way to reason: divide polynomials and use the minimality of the remainder, rather than trying to cancel operator factors. The root result then gives an exact description of the eigenvalues from \(m_T\). These facts concern the eigenvalues and the polynomial identities of an operator; they do not, by themselves, assert that the operator has a basis of eigenvectors.
Check Your Understanding
Use polynomial division, the definition of an eigenvector, and direct matrix calculations to justify each answer.
- If \(p(T)=0\), what degree condition does the remainder have when \(p\) is divided by \(m_T\), and why must that remainder be zero?
- For \(\mathbf{A}=\begin{bmatrix}3&0\\0&-2\end{bmatrix}\), find the minimal polynomial and verify its value at \(\mathbf{A}\).
- Show directly that \(m(x)=x^2\) is the minimal polynomial of \(\begin{bmatrix}0&1\\0&0\end{bmatrix}\).
- Explain where finite-dimensionality is used to prove that every root of \(m_T\) is an eigenvalue.
- Give a concrete reason that a zero product of two matrices does not imply that either factor can be cancelled.
