What the Multiplicity Criterion Says
Diagonalizability asks whether a matrix can be represented using a basis of eigenvectors. When an eigenvalue is repeated, knowing that it occurs several times in the characteristic polynomial does not by itself tell us how many independent eigenvectors it has. The multiplicity criterion makes this distinction precise: for a real matrix whose characteristic polynomial splits over \(\mathbb{R}\), diagonalizability is equivalent to having enough eigenvectors at every eigenvalue to match its algebraic multiplicity.
Let \(\mathbf{A}\in\mathbb{R}^{n\times n}\), with \(n\geq 1\). Its characteristic polynomial is \(p_{\mathbf{A}}(t)=\det(\mathbf{A}-t\mathbf{I}_n)\). It splits over \(\mathbb{R}\) if it factors into linear factors with real roots. For an eigenvalue \(\lambda\), its algebraic multiplicity is the number of times the factor \((\lambda-t)\) occurs in this factorization. Its eigenspace is \(E_\lambda=\{\mathbf{x}\in\mathbb{R}^n:\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\}\), and its geometric multiplicity is \(\dim E_\lambda\). A matrix is diagonalizable over \(\mathbb{R}\) if \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\) for some invertible real matrix \(\mathbf{P}\) and real diagonal matrix \(\mathbf{D}\).
The hypothesis that the characteristic polynomial splits matters: it ensures there are enough real eigenvalues to account for all \(n\) roots, counting multiplicity. The theorem concerns a comparison made separately at each eigenvalue; matching the total number of eigenvectors to \(n\) without checking the eigenspaces would not establish that comparison.
A Bound Needed for the Proof
First, the geometric multiplicity of any eigenvalue is at most its algebraic multiplicity. This bound explains why equality is the relevant test: an eigenspace cannot contribute more independent eigenvectors than the eigenvalue’s characteristic-polynomial multiplicity allows.
Proof. Set \(g=\dim E_\lambda\). Choose a basis of \(E_\lambda\), then extend it to a basis of \(\mathbb{R}^n\). In this basis, the first \(g\) basis vectors are eigenvectors with eigenvalue \(\lambda\). Therefore the matrix representing \(\mathbf{A}\) has block form \(\begin{bmatrix}\lambda\mathbf{I}_g&\mathbf{C}\\\mathbf{O}&\mathbf{B}\end{bmatrix}\), for some matrices \(\mathbf{C}\) and \(\mathbf{B}\). Its characteristic polynomial is \(\det((\lambda-t)\mathbf{I}_g)\det(\mathbf{B}-t\mathbf{I}_{n-g})=(\lambda-t)^g\det(\mathbf{B}-t\mathbf{I}_{n-g})\). Changing basis does not change the characteristic polynomial: the new matrix is \(\mathbf{Q}^{-1}\mathbf{A}\mathbf{Q}\) for an invertible \(\mathbf{Q}\), and determinant multiplicativity gives \(\det(\mathbf{Q}^{-1}(\mathbf{A}-t\mathbf{I}_n)\mathbf{Q})=\det(\mathbf{A}-t\mathbf{I}_n)\). Thus \((\lambda-t)^g\) divides \(p_{\mathbf{A}}(t)\), so \(\operatorname{am}(\lambda)\geq g\). If \(g=n\), the lower-right block has size zero and its determinant is taken to be \(1\); the same argument applies. \(\square\)
Proof of the Criterion
We prove the two implications separately. The first shows that diagonalizability forces each eigenspace to be as large as the characteristic polynomial permits. The second uses all the eigenspaces together to construct a basis.
Proof. Suppose first that \(\mathbf{A}\) is diagonalizable over \(\mathbb{R}\), so \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\). Equivalently, \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\). The columns of \(\mathbf{P}\) are therefore a basis of eigenvectors, and the diagonal entries of \(\mathbf{D}\) are their eigenvalues. For each eigenvalue \(\lambda\), let \(r_\lambda\) be the number of diagonal entries of \(\mathbf{D}\) equal to \(\lambda\). Since \(p_{\mathbf{A}}(t)=\det(\mathbf{D}-t\mathbf{I}_n)\), its factor \((\lambda-t)\) occurs exactly \(r_\lambda\) times. Hence \(r_\lambda=\operatorname{am}(\lambda)\). The \(r_\lambda\) basis vectors associated with \(\lambda\) are independent members of \(E_\lambda\), so \(\dim E_\lambda\geq r_\lambda\). The lemma gives \(\dim E_\lambda\leq\operatorname{am}(\lambda)=r_\lambda\). Together these inequalities give \(\dim E_\lambda=\operatorname{am}(\lambda)\).
Conversely, suppose that \(p_{\mathbf{A}}(t)\) splits over \(\mathbb{R}\), and that \(\dim E_\lambda=\operatorname{am}(\lambda)\) for every eigenvalue. Write the distinct eigenvalues as \(\lambda_1,\dots,\lambda_k\), with algebraic multiplicities \(a_1,\dots,a_k\). The degree of the characteristic polynomial is \(n\), so \(\sum_{i=1}^k a_i=n\). Choose a basis for each \(E_{\lambda_i}\), and combine these lists into one list. To show that this combined list is independent, consider vectors \(\mathbf{x}_i\in E_{\lambda_i}\) satisfying \(\mathbf{x}_1+\cdots+\mathbf{x}_k=\mathbf{0}\). Fix \(i\), and apply the operator \(\prod_{j\ne i}(\mathbf{A}-\lambda_j\mathbf{I}_n)\) to both sides. Every term \(\mathbf{x}_j\) with \(j\ne i\) becomes zero, because one factor is \(\mathbf{A}-\lambda_j\mathbf{I}_n\). The remaining term becomes \(\bigl(\prod_{j\ne i}(\lambda_i-\lambda_j)\bigr)\mathbf{x}_i=\mathbf{0}\). The scalar product is nonzero because the eigenvalues are distinct, so \(\mathbf{x}_i=\mathbf{0}\). If \(k=1\), the product has no factors and means the identity operator, giving the same conclusion directly. This holds for every \(i\), proving independence.
The combined list contains \(\sum_i\dim E_{\lambda_i}=\sum_i a_i=n\) vectors. In \(\mathbb{R}^n\), an independent list of \(n\) vectors is a basis. Every vector in this basis is an eigenvector, so let \(\mathbf{P}\) have these vectors as its columns and let \(\mathbf{D}\) contain their eigenvalues in the corresponding order on its diagonal. Then \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\); since \(\mathbf{P}\) is invertible, \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\). Thus \(\mathbf{A}\) is diagonalizable. \(\square\)
Applying the Multiplicity Test
Worked Example: Equality at Each Eigenvalue
Determine whether \(\mathbf{A}=\begin{bmatrix}2&0&3\\0&2&0\\0&0&5\end{bmatrix}\) is diagonalizable over \(\mathbb{R}\), using the multiplicity criterion.
The matrix is upper triangular, so its characteristic polynomial is the product of its diagonal entries after subtracting \(t\):
Thus the eigenvalues are \(2\), with algebraic multiplicity \(2\), and \(5\), with algebraic multiplicity \(1\). For \(\lambda=2\), solve \((\mathbf{A}-2\mathbf{I}_3)\mathbf{x}=\mathbf{0}\):
So \(E_2=\operatorname{span}\{(1,0,0)^\top,(0,1,0)^\top\}\) and \(\dim E_2=2\). For \(\lambda=5\),
Hence \(E_5=\operatorname{span}\{(1,0,1)^\top\}\) and \(\dim E_5=1\). Both geometric multiplicities equal their corresponding algebraic multiplicities, so the theorem says \(\mathbf{A}\) is diagonalizable.
Check: \(\mathbf{A}(1,0,0)^\top=(2,0,0)^\top\), \(\mathbf{A}(0,1,0)^\top=(0,2,0)^\top\), and \(\mathbf{A}(1,0,1)^\top=(5,0,5)^\top\). These eigenvectors form a basis: the matrix with them as columns is \(\begin{bmatrix}1&0&1\\0&1&0\\0&0&1\end{bmatrix}\), whose inverse is \(\begin{bmatrix}1&0&-1\\0&1&0\\0&0&1\end{bmatrix}\).
Worked Example: A Geometric Multiplicity That Falls Short
Use the criterion to decide whether \(\mathbf{B}=\begin{bmatrix}1&0&0\\0&1&1\\0&0&1\end{bmatrix}\) is diagonalizable over \(\mathbb{R}\).
Since \(\mathbf{B}\) is upper triangular, \(p_{\mathbf{B}}(t)=(1-t)^3\). The sole eigenvalue is \(1\), with algebraic multiplicity \(3\). Its eigenspace is found from
Therefore \(E_1=\operatorname{span}\{(1,0,0)^\top,(0,1,0)^\top\}\), so \(\dim E_1=2\), less than the algebraic multiplicity \(3\). The criterion shows that \(\mathbf{B}\) is not diagonalizable.
Check: \(\mathbf{B}(1,0,0)^\top=(1,0,0)^\top\) and \(\mathbf{B}(0,1,0)^\top=(0,1,0)^\top\), but \(\mathbf{B}(0,0,1)^\top=(0,1,1)^\top\ne(0,0,1)^\top\). The eigenspace calculation accounts for exactly two independent eigenvectors, not three.
Worked Example: Checking the Splitting Hypothesis
Can the real multiplicity criterion be applied to \(\mathbf{C}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}\)?
Compute its characteristic polynomial directly:
For every real \(t\), \(t^2\geq0\), so \(t^2+1>0\). Thus \(p_{\mathbf{C}}(t)\) has no real roots and does not split over \(\mathbb{R}\). The theorem’s hypothesis is absent, so this criterion cannot be used to decide real diagonalizability in this case.
Check: substituting the possible real-root equation \(t^2+1=0\) would require \(t^2=-1\), which no real number satisfies. This confirms that the characteristic polynomial does not split over the stated field.
Reading the Conclusion Correctly
The criterion does not say that a repeated eigenvalue prevents diagonalization. It says to compare two counts for each eigenvalue. Algebraic multiplicity counts how many times the eigenvalue appears in the characteristic polynomial; geometric multiplicity counts the dimension of the space of its eigenvectors. The first example has a repeated eigenvalue and meets the test. The second has a repeated eigenvalue whose eigenspace is too small.
| Comparison for every eigenvalue | Conclusion when the polynomial splits over \(\mathbb{R}\) |
|---|---|
| Geometric multiplicity equals algebraic multiplicity | The matrix is diagonalizable over \(\mathbb{R}\). |
| Geometric multiplicity is smaller for at least one eigenvalue | The matrix is not diagonalizable over \(\mathbb{R}\). |
| The characteristic polynomial does not split over \(\mathbb{R}\) | This real-field criterion does not apply. |
Common mistake: counting repeated roots as separate eigenspaces. For \(\mathbf{B}\), one might say, “The characteristic polynomial has three roots, so there must be three independent eigenvectors.” This confuses the root’s algebraic multiplicity with the dimension of its eigenspace. The calculation gives \((\mathbf{B}-\mathbf{I}_3)\mathbf{x}=\mathbf{0}\) exactly when \(x_3=0\), leaving only two free coordinates. Thus \(\dim E_1=2\), while \(\operatorname{am}(1)=3\); the correction is that the matrix is not diagonalizable.
The proof also explains why checking each eigenspace is sufficient. Distinct eigenspaces contribute independent vectors, and when their dimensions add to \(n\), their eigenvector bases together form a basis of \(\mathbb{R}^n\). Conversely, an eigenbasis can supply only as many independent vectors at a given eigenvalue as its eigenspace contains. The multiplicity criterion turns these observations into an exact test, provided the characteristic polynomial splits over the real field.
Check Your Understanding
Use the definitions and proof above to answer each question.
- What are the algebraic and geometric multiplicities of each eigenvalue in the first worked example?
- For the second example, which equation in \((\mathbf{B}-\mathbf{I}_3)\mathbf{x}=\mathbf{0}\) restricts the eigenspace?
- Why does the proof of the criterion require the characteristic polynomial to split over \(\mathbb{R}\)?
- In the lemma, why does choosing the first basis vectors from \(E_\lambda\) produce a block with \(\lambda\mathbf{I}_g\) in its upper-left corner?
- If each eigenspace dimension equals its algebraic multiplicity and the characteristic polynomial splits, why do the dimensions of all eigenspaces sum to \(n\)?
