Minimal Polynomials and Diagonalizability

Why the Minimal Polynomial Detects Diagonalizability

Diagonalizing an operator means finding a basis in which its action has no interaction between different coordinates: each basis vector is simply scaled by an eigenvalue. The minimal polynomial records which polynomial identities the operator satisfies. The connection is that a diagonal basis allows an annihilating polynomial with one factor for each distinct eigenvalue, while a repeated factor in the minimal polynomial signals that no such basis exists.

Throughout this tutorial, \(V\) is a nonzero finite-dimensional vector space over \(\mathbb{C}\), and \(T:V\to V\) is linear. For a polynomial \(f(t)=a_0+a_1t+\cdots+a_mt^m\), its evaluation at \(T\) is \(f(T)=a_0\operatorname{id}_V+a_1T+\cdots+a_mT^m\). A polynomial \(f\) annihilates \(T\) if \(f(T)\) is the zero map. Existence of a nonzero annihilating polynomial follows from the Cayley–Hamilton theorem: the characteristic polynomial annihilates a matrix of \(T\).

A polynomial acts simply on an eigenvector. This observation gives a useful necessary condition on every annihilating polynomial.

Proof. By the definition of eigenvalue, there is a vector \(\mathbf{v}\ne\mathbf{0}\) with \(T\mathbf{v}=\lambda\mathbf{v}\). Induction gives \(T^j\mathbf{v}=\lambda^j\mathbf{v}\) for each nonnegative integer \(j\). Therefore, if \(f(t)=\sum_{j=0}^m a_jt^j\), then \(f(T)\mathbf{v}=\sum_{j=0}^m a_j\lambda^j\mathbf{v}=f(\lambda)\mathbf{v}\). Since \(f(T)\) is the zero map, \(f(\lambda)\mathbf{v}=\mathbf{0}\). The vector \(\mathbf{v}\) is nonzero, so \(f(\lambda)=0\). \(\square\)

Distinct Eigenvalues Give Independent Directions

The proof of the criterion will use the fact that eigenvectors for different eigenvalues cannot cancel one another. The statement concerns vectors chosen from different eigenspaces; it does not say that any arbitrary collection of vectors from one eigenspace is independent.

Proof. We use induction on \(k\). For \(k=1\), the equation \(\mathbf{v}_1=\mathbf{0}\) already gives the result. Suppose the statement holds for \(k-1\) distinct eigenvalues and that \(\mathbf{v}_1+\cdots+\mathbf{v}_k=\mathbf{0}\). Apply \(T-\lambda_k\operatorname{id}_V\) to both sides. Since \(T\mathbf{v}_i=\lambda_i\mathbf{v}_i\), the resulting equation is \(\sum_{i=1}^{k-1}(\lambda_i-\lambda_k)\mathbf{v}_i=\mathbf{0}\); the term involving \(\mathbf{v}_k\) is zero. The first \(k-1\) vectors still belong to eigenspaces for distinct eigenvalues, so the induction hypothesis gives \((\lambda_i-\lambda_k)\mathbf{v}_i=\mathbf{0}\) for each \(i<k\). Distinctness gives \(\lambda_i-\lambda_k\ne0\), hence \(\mathbf{v}_i=\mathbf{0}\) for \(i<k\). The original sum then gives \(\mathbf{v}_k=\mathbf{0}\). \(\square\)

The Minimal-Polynomial Criterion

Proof. First suppose \(T\) is diagonalizable. Choose a basis of eigenvectors, and list the distinct eigenvalues appearing in that basis as \(\lambda_1,\dots,\lambda_k\). Set \(q(t)=\prod_{i=1}^k(t-\lambda_i)\). On each basis vector \(\mathbf{v}\) with eigenvalue \(\lambda_i\), we have \(q(T)\mathbf{v}=q(\lambda_i)\mathbf{v}=\mathbf{0}\). A linear map that sends every vector of a basis to zero is the zero map, so \(q(T)=\mathbf{0}\). Now let \(f\) be any polynomial that annihilates \(T\). The lemma shows \(f(\lambda_i)=0\) for every \(i\). By the factor theorem, each \(t-\lambda_i\) divides \(f(t)\); because these factors have distinct roots, their product \(q(t)\) divides \(f(t)\). Thus no nonzero annihilating polynomial has degree less than \(k\), and the monic polynomial \(q\) is the minimal polynomial.

Conversely, suppose \(m_T(t)=p(t)=\prod_{i=1}^k(t-\lambda_i)\), with the \(\lambda_i\) distinct. For each \(i\), define the polynomial $$ e_i(t)=\frac{\prod_{j\ne i}(t-\lambda_j)}{\prod_{j\ne i}(\lambda_i-\lambda_j)}. $$ The denominator is nonzero because the roots are distinct. These polynomials satisfy \(e_i(\lambda_j)=1\) if \(i=j\) and \(e_i(\lambda_j)=0\) if \(i\ne j\). The polynomial \(\sum_{i=1}^k e_i(t)-1\) has degree at most \(k-1\) and vanishes at all \(k\) distinct values \(\lambda_1,\dots,\lambda_k\). A nonzero polynomial of degree at most \(k-1\) cannot have \(k\) distinct roots, so \(\sum_i e_i(t)=1\).

For any \(\mathbf{v}\in V\), this identity gives \(\mathbf{v}=\sum_i e_i(T)\mathbf{v}\). Also, $$ (T-\lambda_i\operatorname{id}_V)e_i(T) =\frac{p(T)}{\prod_{j\ne i}(\lambda_i-\lambda_j)} =\mathbf{0}, $$ because \(p(T)=m_T(T)=\mathbf{0}\). Hence each \(e_i(T)\mathbf{v}\) belongs to \(E_{\lambda_i}\), and the eigenspaces together span \(V\). To see that their sum is direct, suppose \(\sum_i\mathbf{w}_i=\mathbf{0}\) with \(\mathbf{w}_i\in E_{\lambda_i}\). Apply \(e_j(T)\) to the equation. Since \(e_j(T)\mathbf{w}_i=e_j(\lambda_i)\mathbf{w}_i\), the result is \(\mathbf{w}_j=\mathbf{0}\). This holds for each \(j\). Choose a basis of each eigenspace; their combined list spans \(V\) and is linearly independent by the direct-sum property. It is therefore an eigenvector basis, so \(T\) is diagonalizable. \(\square\)

Applying the Criterion

Common Pitfall: Confusing the Two Polynomials

A tempting but incorrect calculation is to infer the minimal polynomial from the characteristic polynomial alone. For \(\mathbf{A}=\begin{bmatrix}4&1\\0&4\end{bmatrix}\), the characteristic polynomial is \(\det(\mathbf{A}-t\mathbf{I}_2)=(4-t)^2\). One might wrongly conclude that the minimal polynomial is \(t-4\), because \(4\) is the only eigenvalue. The correction is to test polynomial evaluation: \(\mathbf{A}-4\mathbf{I}_2=\begin{bmatrix}0&1\\0&0\end{bmatrix}\ne\mathbf{O}\), while its square is \(\mathbf{O}\). So the minimal polynomial is \((t-4)^2\), and the matrix is not diagonalizable.

The reverse mistake is to treat a repeated root of the characteristic polynomial as proof of non-diagonalizability. In the third example, the characteristic polynomial has the repeated root \(1\), but the minimal polynomial has no repeated factor and the matrix is diagonalizable. The criterion concerns the minimal polynomial: its factors describe the shortest polynomial identity satisfied by the operator.

The field matters as well. The theorem above is for complex vector spaces and complex polynomials. A real matrix may have a minimal polynomial that does not split into real linear factors; that prevents diagonalization over \(\mathbb{R}\), but does not rule out diagonalization after allowing complex scalars. Always identify the scalar field before applying the criterion.