Algebraic Multiplicity of a Generalized Eigenspace

Why a Generalized Eigenspace Has the “Right” Dimension

An eigenvalue can occur more times in the characteristic polynomial than there are linearly independent eigenvectors for it. Generalized eigenvectors account for the additional dimensions: they form a larger invariant subspace on which the operator differs from scalar multiplication by a nilpotent map. The precise connection is that the dimension of this generalized eigenspace equals the algebraic multiplicity of the eigenvalue.

Work over \(\mathbb{C}\), and let \(V\) be a finite-dimensional vector space with \(\dim V=n\). For a linear map \(T:V\to V\), its characteristic polynomial in any basis is \(p_T(t)=\det([T]-t\mathbf{I}_n)\). This polynomial does not depend on the chosen basis: if a change of basis gives the matrix \(\mathbf{P}^{-1}[T]\mathbf{P}\), then \([T]-t\mathbf{I}_n\) changes by the same similarity, which preserves the determinant. The algebraic multiplicity of \(\lambda\) is the exponent of the factor \(t-\lambda\) in \(p_T(t)\); using \(\lambda-t\) instead gives the same exponent.

The key input from primary decomposition is that, over \(\mathbb{C}\), \(V\) is the direct sum of its generalized eigenspaces, one for each distinct eigenvalue of \(T\). Each summand is invariant under \(T\), meaning that \(T\) maps it into itself. We will use this decomposition, but the dimension claim requires an additional argument: on a given summand, the restriction of \(T\) is a scalar map plus a nilpotent map.

The Characteristic Polynomial of a Nilpotent Map

A linear map \(N:U\to U\) is nilpotent if \(N^k=\mathbf{0}\) for some positive integer \(k\). We first establish the basis property that lets us compute its characteristic polynomial. Proving the stronger basis statement is useful: the determinant formula then follows directly from the diagonal entries, rather than from an unproved assumption about a quotient.

Proof. We prove the basis statement by induction on \(d=\dim U\). If \(d=0\), the empty basis has the required property. Suppose \(d\geq 1\). Since \(N\) is nilpotent, choose \(k\geq1\) with \(N^k=\mathbf{0}\). Its kernel contains a nonzero vector: if \(N=\mathbf{0}\), any nonzero vector is in the kernel; otherwise choose the least such \(k\), take \(x\) with \(N^{k-1}x\ne\mathbf{0}\), and set \(v_1=N^{k-1}x\). In either case \(Nv_1=\mathbf{0}\). The one-dimensional space \(L=\operatorname{span}\{v_1\}\) is invariant under \(N\), so \(N\) induces a nilpotent map \(\overline{N}\) on the quotient space \(U/L\). That quotient has dimension \(d-1\). By the induction hypothesis, it has a basis in which \(\overline{N}\) is strictly upper triangular. Lift those basis vectors to \(v_2,\dots,v_d\in U\). Then \(v_1,\dots,v_d\) form a basis of \(U\). The first column of the matrix of \(N\) in this basis is zero because \(Nv_1=\mathbf{0}\), and the lower-right \((d-1)\times(d-1)\) block is the matrix of \(\overline{N}\), which is strictly upper triangular. The first row may have entries to the right of its first entry, but those entries are above the diagonal. Thus the full matrix is strictly upper triangular. Its diagonal entries are all zero, so \([N]-t\mathbf{I}_d\) is upper triangular with every diagonal entry \(-t\). Its determinant is the product of those entries, namely \((-t)^d\). \(\square\)

Now let \(T\) act on a \(d\)-dimensional invariant subspace \(U\), and suppose \(T|_U=\lambda\operatorname{id}_U+N\) for a nilpotent \(N\). In the basis supplied by the lemma, the matrix of \(T|_U-t\operatorname{id}_U\) is upper triangular with diagonal entries \(\lambda-t\). Therefore its characteristic polynomial is \((\lambda-t)^d\). This calculation is the local reason a summand of dimension \(d\) contributes exactly \(d\) copies of the eigenvalue \(\lambda\).

The Dimension and Algebraic Multiplicity Theorem

Proof. Let \(\lambda_1,\dots,\lambda_r\) be the distinct eigenvalues of \(T\). By primary decomposition, \(V=G_{\lambda_1}\oplus\cdots\oplus G_{\lambda_r}\), and every summand is invariant under \(T\). Set \(d_i=\dim G_{\lambda_i}\). On \(G_{\lambda_i}\), the restriction of \(T-\lambda_i\operatorname{id}_V\) is nilpotent, since its \(n\)th power is zero there by the definition of \(G_{\lambda_i}\). The nilpotent lemma therefore gives the characteristic polynomial \((\lambda_i-t)^{d_i}\) for the restriction of \(T\) to that summand. Choose a basis for each summand and concatenate these bases. In the resulting basis for \(V\), the matrix of \(T\) is block diagonal with one block for each restriction. The determinant of a block diagonal matrix is the product of the determinants of its diagonal blocks, so $$ p_T(t)=\prod_{i=1}^{r}(\lambda_i-t)^{d_i}. $$ The \(\lambda_i\) are distinct, so the factor belonging to \(\lambda_i\) occurs with exponent exactly \(d_i\), with no contribution from any other factor. Thus \(\operatorname{am}(\lambda_i)=d_i=\dim G_{\lambda_i}\) for every \(i\), including the specified \(\lambda\). \(\square\)

The theorem identifies the size of each primary summand without requiring a basis of Jordan chains. The proof uses only the direct-sum decomposition, invariance of its summands, and the characteristic polynomial of a scalar-plus-nilpotent restriction. In particular, it does not say that the eigenspace itself has dimension equal to the algebraic multiplicity: the eigenspace is \(\ker(T-\lambda\operatorname{id}_V)\), which can be smaller than \(G_\lambda\).

Worked Checks

Common Pitfall: Confusing Algebraic and Geometric Multiplicity

The geometric multiplicity of \(\lambda\) is the dimension of the ordinary eigenspace \(\ker(T-\lambda\operatorname{id}_V)\). It counts independent eigenvectors, not all generalized eigenvectors. For example, take \(\mathbf{B}=\begin{bmatrix}2&1\\0&2\end{bmatrix}\). A mistaken calculation might find \(\ker(\mathbf{B}-2\mathbf{I}_2)=\operatorname{span}\{(1,0)^{\top}\}\), conclude that its dimension \(1\) is the algebraic multiplicity, and stop. The first kernel calculation is correct, but that conclusion is not: \(p_{\mathbf{B}}(t)=(2-t)^2\), so the algebraic multiplicity is \(2\).

The correction is to compute the generalized eigenspace using the power prescribed by the definition. Here \(\mathbf{B}-2\mathbf{I}_2=\begin{bmatrix}0&1\\0&0\end{bmatrix}\), whose square is \(\mathbf{O}\). Thus \(\ker(\mathbf{B}-2\mathbf{I}_2)^2=\mathbb{C}^2\), of dimension \(2\). Multiplying back confirms that \((\mathbf{B}-2\mathbf{I}_2)^2\mathbf{x}=\mathbf{0}\) for every \(\mathbf{x}\in\mathbb{C}^2\). The ordinary eigenspace has dimension \(1\), while the generalized eigenspace has dimension \(2\), in agreement with the theorem.

How to Use the Result

When a characteristic polynomial has a factor \((\lambda-t)^m\), the theorem tells you that the corresponding generalized eigenspace has dimension \(m\). Conversely, once primary decomposition is known, the dimensions of its summands determine the exponents in the characteristic polynomial. This gives a reliable way to check calculations, especially when an ordinary eigenspace is too small to account for the full algebraic multiplicity.

The argument depends on finite dimension and on working over a field where the characteristic polynomial splits into linear factors. Over \(\mathbb{C}\), every characteristic polynomial splits. For a real operator with nonreal eigenvalues, the same statement is applied after extending the scalars to \(\mathbb{C}\); nonreal eigenvalues occur together with their complex conjugates. In every case, keep the distinction in view: the theorem equates algebraic multiplicity with the dimension of a generalized eigenspace, not necessarily with the number of independent ordinary eigenvectors.