Why Geometric Multiplicity Cannot Exceed Algebraic Multiplicity

Two Ways to Measure an Eigenvalue

An eigenvalue can be counted in two different ways. Its eigenspace records how many linearly independent directions are eigenvector directions. Its characteristic polynomial records how many times the eigenvalue occurs as a root. These counts need not be equal, but one cannot exceed the other. The reason is structural: putting a basis of the eigenspace first in a basis for the whole space forces a block form, and that block form forces a corresponding factor in the characteristic polynomial.

Let \(\mathbb{F}\) be \(\mathbb{R}\) or \(\mathbb{C}\), and let \(\mathbf{A}\in\mathbb{F}^{n\times n}\). For this tutorial, take an eigenvalue \(\lambda\in\mathbb{F}\). Its eigenspace is \(E_{\lambda}=\{\mathbf{x}\in\mathbb{F}^n:\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\}\). An eigenvalue has a nonzero eigenvector, so its eigenspace has positive dimension. The characteristic polynomial, with the course convention, is \(p_{\mathbf{A}}(t)=\det(\mathbf{A}-t\mathbf{I}_n)\).

The eigenspace is a subspace of the finite-dimensional space \(\mathbb{F}^n\), so it has a basis. Write \(k=\dim E_{\lambda}\), and choose a basis \(\mathbf{v}_1,\dots,\mathbf{v}_k\) for it. Extend that list to a basis \(\mathbf{v}_1,\dots,\mathbf{v}_k,\mathbf{w}_{k+1},\dots,\mathbf{w}_n\) for \(\mathbb{F}^n\). This extension is possible by appending vectors from any spanning list for \(\mathbb{F}^n\) whenever they are not already in the span; when the list spans the space, the resulting independent list is a basis.

What the Adapted Basis Reveals

Let \(\mathbf{P}\) be the matrix whose columns are this extended basis. The matrix of the same linear map in the adapted basis is \(\mathbf{B}=\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\). For each \(i\leq k\), the eigenvector equation gives \(\mathbf{A}\mathbf{v}_i=\lambda\mathbf{v}_i\). Thus the first \(k\) columns of \(\mathbf{B}\) have \(\lambda\) in their corresponding positions and zeros elsewhere. In block form, this says

where the lower-left block is zero. The block \(\mathbf{D}\) need not be zero: the remaining basis vectors may be mapped partly into the eigenspace. What matters for the argument is that the eigenvector basis vectors themselves map into the eigenspace, which makes the lower-left block zero.

Changing basis does not change the characteristic polynomial. In fact, \(\mathbf{B}-t\mathbf{I}_n=\mathbf{P}^{-1}(\mathbf{A}-t\mathbf{I}_n)\mathbf{P}\). By the determinant product rule \(\det(\mathbf{X}\mathbf{Y})=\det(\mathbf{X})\det(\mathbf{Y})\), and \(\det(\mathbf{P}^{-1})\det(\mathbf{P})=1\), so \(\det(\mathbf{B}-t\mathbf{I}_n)=\det(\mathbf{A}-t\mathbf{I}_n)\).

Proof. Set \(k=\operatorname{gm}(\lambda)\), and use the adapted basis above. Since \(\lambda\) is an eigenvalue, \(k\geq1\). If \(k=n\), every basis vector is an eigenvector, so \(\mathbf{B}=\lambda\mathbf{I}_n\) and \(p_{\mathbf{A}}(t)=\det(\lambda\mathbf{I}_n-t\mathbf{I}_n)=(\lambda-t)^n\). Hence \(\operatorname{am}(\lambda)=n=k\).

Now suppose \(k<n\). Then \(\mathbf{B}-t\mathbf{I}_n\) is block upper triangular:

For a block upper triangular matrix \(\begin{bmatrix}\mathbf{X}&\mathbf{Y}\\\mathbf{O}&\mathbf{Z}\end{bmatrix}\), its determinant is \(\det(\mathbf{X})\det(\mathbf{Z})\). To see why, use the permutation expansion of the determinant: a nonzero term must assign the rows of the lower block to columns of the right block, since the entries in the lower-left block are zero. The remaining rows must then use columns of the left block. The surviving terms split into the determinant expansions of \(\mathbf{X}\) and \(\mathbf{Z}\), with their signs and products, giving the stated product. Applying this fact and the equality of characteristic polynomials under change of basis gives

Since \((\lambda-t)^k=(-1)^k(t-\lambda)^k\), the polynomial \(p_{\mathbf{A}}(t)\) is divisible by \((t-\lambda)^k\). Therefore \(\operatorname{am}(\lambda)\geq k=\operatorname{gm}(\lambda)\). Finally, \(p_{\mathbf{A}}\) has degree \(n\), so no root can have multiplicity greater than \(n\). This proves all the claimed inequalities. \(\square\)

Worked Examples

Why the Inequality Matters

The inequality is a constraint, not a claim that the two multiplicities are always equal. A root of multiplicity \(r\) has room for at most \(r\) independent eigenvector directions. If the eigenspace has dimension \(r\), the bound is sharp; if its dimension is smaller, the characteristic polynomial counts more copies of the root than there are independent eigenvectors. The adapted-basis argument identifies the source of the bound: each basis vector in the eigenspace contributes a diagonal \(\lambda\) block, and those \(k\) entries force at least \(k\) copies of the root.

Common mistake: assuming a repeated root automatically gives many independent eigenvectors. For the first example, the characteristic polynomial is \((t-2)^2\). A mistaken calculation might conclude that there must be two independent eigenvectors for \(\lambda=2\), and therefore set \(\operatorname{gm}(2)=2\). But solving the eigenvector equation gives \(x_2=0\), so every eigenvector is a multiple of \(\begin{bmatrix}1\\0\end{bmatrix}\), and \(\operatorname{gm}(2)=1\). The correct conclusion is \(1\leq2\), not equality. Root multiplicity counts a polynomial factor; geometric multiplicity counts independent vectors.

This comparison is a useful diagnostic whenever a characteristic polynomial has a repeated root. Compute the dimension of the eigenspace independently from the characteristic polynomial, then check that it is positive and no larger than the root multiplicity. The inequality itself does not decide whether there are enough eigenvectors to form a basis; it establishes the necessary bound without claiming more.