Repeated Eigenvalues And Eigenspace Dimensions

One Repeated Root, Different Eigenspaces

A repeated root of a characteristic polynomial tells us how often an eigenvalue occurs as a root, but it does not by itself say how many independent directions of eigenvectors belong to that eigenvalue. For instance, two matrices can both have characteristic polynomial \((2-t)^2\), yet one can have a one-dimensional eigenspace and the other a two-dimensional eigenspace. To distinguish these situations, we compare algebraic multiplicity with geometric multiplicity.

Throughout, matrices and vectors are over \(\mathbb{R}\). For an \(n\times n\) matrix \(\mathbf{A}\), the characteristic polynomial is \(p_{\mathbf{A}}(t)=\det(\mathbf{A}-t\mathbf{I}_n)\). An eigenvalue \(\lambda\) is a scalar for which there is a nonzero vector \(\mathbf{v}\) satisfying \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). The eigenspace for \(\lambda\) consists of all such vectors together with the zero vector.

These are different measurements. Algebraic multiplicity is read from a polynomial; geometric multiplicity is found by solving a vector equation and measuring the dimension of its solution space. In particular, geometric multiplicity is not the number of eigenvectors: every nonzero scalar multiple of an eigenvector is another eigenvector, so an eigenspace of positive dimension contains infinitely many eigenvectors.

From the Eigenvector Equation to a Dimension

Rearranging \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\) gives \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{v}=\mathbf{0}\). Thus an eigenspace is a null space, and its dimension can be found by solving a homogeneous system. The rank of a matrix is the number of pivot columns in its reduced row echelon form; the nullity is the dimension of its null space.

Proof. A vector \(\mathbf{v}\) belongs to \(E_{\lambda}\) exactly when \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). Subtracting \(\lambda\mathbf{v}\) from both sides shows this is equivalent to \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{v}=\mathbf{0}\). Therefore the set of vectors in \(E_{\lambda}\) is exactly the null space of \(\mathbf{A}-\lambda\mathbf{I}_n\). By the Rank-Nullity Theorem, the dimension of the null space of an \(n\)-column matrix is \(n\) minus its rank. Since geometric multiplicity is the dimension of \(E_{\lambda}\), the stated formula follows. \(\square\)

This gives a practical procedure: first find the repeated root and its algebraic multiplicity from \(p_{\mathbf{A}}(t)\); then substitute that eigenvalue into \(\mathbf{A}-\lambda\mathbf{I}_n\), solve the resulting homogeneous system, and count the free variables. The count of free variables is the geometric multiplicity.

Same Algebraic Multiplicity, Smaller Eigenspace

The calculation shows why the repeated root does not settle the dimension question. The off-diagonal entry in this matrix imposes the condition \(x_2=0\), leaving only one free direction in the eigenspace.

Same Algebraic Multiplicity, Larger Eigenspace

Proof. Subtracting \(t\mathbf{I}_n\) gives \(\mathbf{A}-t\mathbf{I}_n=(\lambda-t)\mathbf{I}_n\), a diagonal matrix with \(n\) diagonal entries equal to \(\lambda-t\). Its determinant is therefore \((\lambda-t)^n\), so the algebraic multiplicity of \(\lambda\) is \(n\). Also, \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\) for every \(\mathbf{v}\in\mathbb{R}^n\), so \(E_{\lambda}=\mathbb{R}^n\) and its dimension is \(n\). \(\square\)

A Three-Dimensional Example

Here a root of multiplicity \(3\) corresponds to an eigenspace of dimension \(2\). Together with the earlier examples, this illustrates three distinct outcomes: algebraic multiplicity \(2\) with geometric multiplicity \(1\), algebraic multiplicity \(2\) with geometric multiplicity \(2\), and algebraic multiplicity \(3\) with geometric multiplicity \(2\). The computations of the two multiplicities answer different questions, so both are needed to describe a repeated eigenvalue.

Common Mistake: Equating the Two Multiplicities

A tempting but incorrect calculation is: “The characteristic polynomial is \((2-t)^2\), so the eigenspace has dimension \(2\).” This infers geometric multiplicity directly from algebraic multiplicity, without solving the eigenvector equation. For \(\mathbf{A}=\begin{bmatrix}2&1\\0&2\end{bmatrix}\), that claim would give \(\operatorname{gm}(2)=2\), but the actual system is

There is only one free variable, so the correction is \(\operatorname{gm}(2)=1\), whereas \(\operatorname{am}(2)=2\). The reliable approach is to compute the characteristic polynomial and then independently solve \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{v}=\mathbf{0}\). The earlier result that geometric multiplicity cannot exceed algebraic multiplicity gives a useful check, but it does not replace the eigenspace calculation.

The comparison matters because repeated roots can conceal different amounts of independent eigenvector information. The characteristic polynomial records root multiplicity; the null space records the available eigenvector directions. Keeping these calculations separate prevents a repeated root from being mistaken for a guarantee of multiple independent eigenvectors.