Primary Decomposition into Generalized Eigenspaces

Why Generalized Eigenspaces Split the Space

An operator need not have enough ordinary eigenvectors to give a basis. Generalized eigenvectors extend the idea of eigenvectors by including vectors that become eigenvectors after repeated application of \(T-\lambda I_V\). The primary decomposition theorem says that, over \(\mathbb{C}\), these generalized eigenspaces account for the entire finite-dimensional space, and each vector has a unique expression as a sum of components from them.

Let \(V\) be a finite-dimensional complex vector space, let \(T:V\to V\) be linear, and write \(n=\dim V\). The identity map on \(V\) is \(I_V\). For a polynomial \(f(x)=c_0+c_1x+\cdots+c_kx^k\), define \(f(T)=c_0I_V+c_1T+\cdots+c_kT^k\). The characteristic polynomial \(p_T(x)\) is \(\det([T]_{\mathcal B}-xI_n)\) in any basis \(\mathcal B\) of \(V\). Its roots are the eigenvalues of \(T\).

The exponent \(n\) is a convenient uniform choice: it allows us to define all generalized eigenspaces without first finding a smallest exponent for each eigenvalue. If \(V=\{\mathbf{0}\}\), there are no eigenvalues and the direct sum over the empty collection is \(\{\mathbf{0}\}\). We consider \(n\geq1\) below.

Bézout Identities Build Polynomial Components

The key algebraic fact is that powers of distinct linear factors are relatively prime. Two polynomials are relatively prime when their only common divisors are nonzero constants. The polynomial Euclidean algorithm then gives a Bézout identity: if \(f\) and \(g\) are relatively prime, there are polynomials \(u\) and \(v\) such that \(uf+vg=1\). Applying this identity to \(T\) turns scalar polynomial relationships into operators on \(V\).

Proof. For each \(i\), let \(r_i=Q/q_i\). Pairwise relative primality implies that \(r_i\) and \(q_i\) are relatively prime: none of the factors in \(r_i\) shares a nonconstant divisor with \(q_i\). By the polynomial Euclidean algorithm, choose \(u_i,v_i\) such that \(u_ir_i+v_iq_i=1\), and set \(e_i=u_ir_i\). Then \(e_i-1=-v_iq_i\), so \(e_i\equiv1\) modulo \(q_i\). If \(j\ne i\), \(r_i\) contains the factor \(q_j\), so \(e_i\equiv0\) modulo \(q_j\). Consequently \(\sum_i e_i-1\) is divisible by every \(q_j\). Since the \(q_j\) are pairwise relatively prime, their product \(Q\) divides this sum minus \(1\). Write \(\sum_i e_i-1=hQ\). Evaluating at \(T\) gives \(\sum_i e_i(T)=I_V\), because \(Q(T)=0\). Also, \(q_i e_i=u_iQ\), so \(q_i(T)e_i(T)=0\) and \(e_i(T)V\subseteq\ker q_i(T)\). Finally, if \(q_j(T)\mathbf{w}_j=0\), the congruences just established imply \(e_i(T)\mathbf{w}_j=\mathbf{w}_j\) for \(i=j\), and \(e_i(T)\mathbf{w}_j=\mathbf{0}\) otherwise. \(\square\)

The lemma gives more than a way to write vectors as sums: its operators isolate one component at a time. The congruences are the essential point. The polynomial \(e_i\) acts like \(1\) on the part annihilated by \(q_i(T)\), and like \(0\) on the parts annihilated by the other factors.

The Primary Decomposition Theorem

Proof. The characteristic polynomial factors over \(\mathbb{C}\) as \(p_T(x)=c\prod_{i=1}^s(x-\lambda_i)^{a_i}\), where \(c\ne0\) and each \(a_i\geq1\). The roots are exactly the eigenvalues: \(p_T(\lambda)=0\) means a matrix for \(T-\lambda I_V\) has determinant zero, hence has a nonzero kernel vector. Since \(\sum_i a_i=n\), in particular \(a_i\leq n\). Define \(q_i(x)=(x-\lambda_i)^n\) and \(Q(x)=\prod_iq_i(x)\). Then the factors \(q_i\) are pairwise relatively prime, and \(p_T\) divides \(Q\), up to the nonzero constant \(c\). The Cayley–Hamilton theorem gives \(p_T(T)=0\), so \(Q(T)=0\) as well. Apply the lemma to obtain polynomials \(e_i\) with \(\sum_i e_i(T)=I_V\) and \(e_i(T)V\subseteq\ker q_i(T)=E_{\lambda_i}\). For any \(\mathbf{v}\in V\), this gives \(\mathbf{v}=\sum_i e_i(T)\mathbf{v}\), proving that the generalized eigenspaces span \(V\). To prove uniqueness, suppose \(\sum_j\mathbf{w}_j=\mathbf{0}\), where \(\mathbf{w}_j\in E_{\lambda_j}=\ker q_j(T)\). Apply \(e_i(T)\) to this equation. By the lemma, all terms except \(\mathbf{w}_i\) become zero, and \(e_i(T)\mathbf{w}_i=\mathbf{w}_i\). Hence \(\mathbf{w}_i=\mathbf{0}\). This holds for every \(i\), so the sum is direct. \(\square\)

The polynomial \(Q\) in this proof uses the same exponent \(n\) for every eigenvalue. It is chosen to match our definition of \(E_\lambda\); it need not be the smallest polynomial that annihilates \(T\). The proof relies on finite dimensionality both in the exponent and in the characteristic polynomial factorization.

Worked Decompositions

Common Pitfall: Using Only Ordinary Eigenvectors

A frequent error is to use \(\ker(T-\lambda I_V)\) in place of \(E_\lambda\). That only finds ordinary eigenvectors and can miss generalized eigenvectors. In the second example, \((\mathbf{A}-2\mathbf{I}_3)\begin{bmatrix}0\\1\\0\end{bmatrix}=\begin{bmatrix}1\\0\\0\end{bmatrix}\ne\mathbf{0}\). A calculation that concludes \(\begin{bmatrix}0\\1\\0\end{bmatrix}\notin E_2\) from this nonzero result is wrong: the definition tests the third power, and

Indeed, \((\mathbf{A}-2\mathbf{I}_3)^3\) also sends this vector to zero. The ordinary eigenspace for \(2\) is only the span of \(\begin{bmatrix}1\\0\\0\end{bmatrix}\), while the generalized eigenspace also contains \(\begin{bmatrix}0\\1\\0\end{bmatrix}\). This distinction is precisely why the theorem uses generalized eigenspaces.

The theorem is useful because it separates the action of an operator according to its distinct eigenvalues. Each component lies in a subspace associated with one eigenvalue, and the decomposition is unique; computations can therefore be carried out component by component. The polynomial operators in the proof also explain the mechanism: Bézout identities construct operators that select one component and discard the others.