Algebraic and Geometric Multiplicities in the Diagonalization Test

Two Ways to Measure an Eigenvalue

An eigenvalue can be counted in two different ways. The characteristic polynomial records how often it occurs as a root. The eigenspace records how many independent eigenvectors belong to it. These numbers describe related but distinct information: a repeated root does not, by itself, guarantee that the eigenspace has the same dimension as the root’s multiplicity.

Work over \(\mathbb{C}\) and let \(\mathbf{A}\in\mathbb{C}^{n\times n}\), where \(n\) is finite. Its characteristic polynomial is \(p_{\mathbf{A}}(t)=\det(\mathbf{A}-t\mathbf{I}_n)\). A number \(\lambda\in\mathbb{C}\) is an eigenvalue when there is a nonzero vector \(\mathbf{v}\) such that \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). The corresponding eigenspace is \(E_\lambda=\{\mathbf{v}\in\mathbb{C}^n:\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\}\). It includes the zero vector and is a subspace.

The sign convention \(p_{\mathbf{A}}(t)=\det(\mathbf{A}-t\mathbf{I}_n)\) does not change root multiplicities: the factors \((\lambda-t)^k\) and \((t-\lambda)^k\) differ only by a nonzero constant. Since \(\lambda\) is an eigenvalue, \(E_\lambda\) contains a nonzero vector, so its dimension is at least one. Its dimension is a count of independent eigenvectors, not a count of how many vectors the eigenspace contains; any nonzero eigenspace has infinitely many vectors over \(\mathbb{C}\).

Computing Both Multiplicities

These examples show why the multiplicities must not be confused. The characteristic polynomial counts a root with its algebraic multiplicity, while solving \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\) determines the geometric multiplicity. The two numbers can agree, as for \(\lambda=2\) in the first example, or differ, as for \(\lambda=3\) in the second.

Why Geometric Multiplicity Cannot Be Larger

Proof. Because \(\lambda\) is an eigenvalue, there is a nonzero vector in \(E_\lambda\). Thus \(\operatorname{gm}(\lambda)\geq1\). Write \(g=\dim E_\lambda\), choose a basis of \(E_\lambda\), and extend it to a basis of \(\mathbb{C}^n\). Such an extension can be formed by adding standard basis vectors whenever they are not already in the span; after finitely many additions, the list spans \(\mathbb{C}^n\) and remains linearly independent. In this basis the matrix of the linear map represented by \(\mathbf{A}\) has the block form \(\begin{bmatrix}\lambda\mathbf{I}_g&\mathbf{C}\\\mathbf{O}&\mathbf{D}\end{bmatrix}\). Indeed, the first \(g\) basis vectors are in \(E_\lambda\), so each is mapped to \(\lambda\) times itself, and consequently the lower-left block is zero.

The characteristic polynomial is unchanged by a change of basis: if \(\mathbf{S}\) is the basis matrix, the new matrix is \(\mathbf{S}^{-1}\mathbf{A}\mathbf{S}\), and \(\det(\mathbf{S}^{-1}(\mathbf{A}-t\mathbf{I}_n)\mathbf{S})=\det(\mathbf{A}-t\mathbf{I}_n)\) by determinant multiplicativity. Therefore we may compute \(p_{\mathbf{A}}(t)\) from the block matrix. For a block upper-triangular matrix, the determinant is the product of the determinants of its diagonal blocks: in the determinant’s permutation expansion, any nonzero term must match the first \(g\) rows to the first \(g\) columns, leaving the remaining rows and columns matched within the lower-right block. It follows that \(\displaystyle p_{\mathbf{A}}(t)=(\lambda-t)^g\det(\mathbf{D}-t\mathbf{I}_{n-g})\).

If \(g=n\), then every basis vector lies in \(E_\lambda\), so \(\mathbf{A}=\lambda\mathbf{I}_n\) in this basis and \(p_{\mathbf{A}}(t)=(\lambda-t)^n\); the same multiplicity conclusion holds. In either case \((t-\lambda)^g\) divides \(p_{\mathbf{A}}(t)\), since \((\lambda-t)^g=(-1)^g(t-\lambda)^g\). By the definition of algebraic multiplicity, \(\operatorname{am}(\lambda)\geq g=\operatorname{gm}(\lambda)\). \(\square\)

The proof explains the bound: every independent eigenvector for \(\lambda\) supplies one basis direction on which the matrix acts as multiplication by \(\lambda\). Those directions force at least that many factors of \(t-\lambda\) in the characteristic polynomial. The remaining block can contribute still more factors, which is why the inequality need not be an equality.

A Three-Dimensional Comparison

How the Comparison Informs Diagonalization

For a finite-dimensional space, the eigenspace dimension criterion says that a matrix is diagonalizable when its eigenspaces together provide a basis. In particular, the geometric multiplicities measure how many independent eigenvectors each eigenvalue contributes. Algebraic multiplicities give a separate count from the characteristic polynomial, and the theorem bounds each eigenspace’s contribution by its corresponding root multiplicity.

For example, in the three-dimensional worked example, the two eigenspaces have dimensions \(1\) and \(1\), so together they provide only two independent eigenvectors. The matrix acts on a three-dimensional space, so those eigenspaces alone cannot supply a basis. This conclusion comes from counting eigenspace dimensions; it does not come merely from seeing that the characteristic polynomial has a repeated root.

A diagonal matrix offers a useful case where the two counts can be read directly. If \(\mathbf{D}=\operatorname{diag}(d_1,\dots,d_n)\) and \(\lambda\) appears \(r\) times among the diagonal entries, then both multiplicities equal \(r\).

Proof. The characteristic polynomial is \(\prod_{j=1}^{n}(d_j-t)\). Each diagonal entry equal to \(\lambda\) contributes one factor \(\lambda-t\), and each entry not equal to \(\lambda\) contributes a factor nonzero at \(t=\lambda\). Thus the root multiplicity is \(r\). For \(\mathbf{x}=(x_1,\dots,x_n)^\top\), the equation \((\mathbf{D}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\) says \((d_j-\lambda)x_j=0\) for every \(j\). If \(d_j\ne\lambda\), this forces \(x_j=0\); if \(d_j=\lambda\), it places no restriction on \(x_j\). Therefore the eigenspace has one freely chosen coordinate for each of the \(r\) matching diagonal entries, and its dimension is \(r\). \(\square\)

Common Mistake: Counting Roots as Eigenvectors

A repeated root is not automatically a guarantee of the same number of independent eigenvectors. For \(\mathbf{B}=\begin{bmatrix}3&1\\0&3\end{bmatrix}\), the characteristic polynomial is \((3-t)^2\). A wrong calculation might say, “The root \(3\) occurs twice, so the eigenspace has dimension two.” That conclusion treats algebraic multiplicity as though it were geometric multiplicity.

The correction is to solve the eigenspace equation. Here \(\begin{bmatrix}0&1\\0&0\end{bmatrix}(x,y)^\top=(y,0)^\top=\mathbf{0}\), so \(y=0\) and \(x\) is free. The eigenspace is the one-dimensional span of \((1,0)^\top\). Thus \(\operatorname{am}(3)=2\) but \(\operatorname{gm}(3)=1\), in agreement with the proved inequality.

When testing diagonalization, keep the two computations distinct: factor the characteristic polynomial to find algebraic multiplicities, and solve the homogeneous systems \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\) to find geometric multiplicities. A repeated root signals that a comparison is needed; it does not settle the eigenspace dimension.