Two Different Counts
When an eigenvalue is repeated in a characteristic polynomial, it is tempting to interpret that repetition as a count of how many independent eigenvectors the matrix has for that eigenvalue. Those are different counts. The characteristic polynomial records how often an eigenvalue occurs as a root; an eigenspace records all vectors that satisfy the eigenvector equation. Confusing these quantities can lead directly to a wrong conclusion about diagonalizability.
Throughout, let \(\mathbf{A}\in\mathbb{R}^{n\times n}\). Write its characteristic polynomial as \(p_{\mathbf{A}}(t)=\det(\mathbf{A}-t\mathbf{I}_n)\), where \(t\) is a polynomial variable. A real number \(\lambda\) is an eigenvalue if there is a nonzero vector \(\mathbf{v}\) with \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). Its eigenspace is \(E_{\lambda}=\operatorname{Nul}(\mathbf{A}-\lambda\mathbf{I}_n)\), including the zero vector.
The sign convention in \(p_{\mathbf{A}}(t)=\det(\mathbf{A}-t\mathbf{I}_n)\) does not affect the multiplicity: the factors \(t-\lambda\) and \(\lambda-t\) differ only by a nonzero constant. For instance, if the characteristic polynomial contains \((t-3)^2\) but not \((t-3)^3\), then \(3\) has algebraic multiplicity \(2\). If it contains \((t-3)^3\), its algebraic multiplicity is at least \(3\), not \(2\).
Why the Counts Are Related, but Not Equal
For every eigenvalue, the geometric multiplicity is at least \(1\), since its eigenspace contains a nonzero eigenvector. More importantly, the geometric multiplicity can never exceed the algebraic multiplicity. Here is a proof that also explains why the two counts measure different things.
Proof. The lower bound holds because an eigenvalue has a nonzero eigenvector, so \(E_{\lambda}\) has positive dimension. Put \(g=\dim E_{\lambda}\), and choose a basis of \(E_{\lambda}\). Extend it to a basis of \(\mathbb{R}^n\), which is possible in a finite-dimensional vector space. In this basis, the first \(g\) basis vectors are eigenvectors with eigenvalue \(\lambda\). Therefore the matrix representing \(\mathbf{A}\) has block form \(\begin{bmatrix}\lambda\mathbf{I}_g&\mathbf{C}\\\mathbf{O}&\mathbf{D}\end{bmatrix}\) for some blocks \(\mathbf{C}\) and \(\mathbf{D}\). The lower-left block is zero because \(\mathbf{A}\) maps each of the first \(g\) basis vectors to \(\lambda\) times itself.
This block triangular determinant is the characteristic polynomial in the chosen basis. Changing basis does not change the characteristic polynomial: if \(\mathbf{S}\) is the change-of-basis matrix, then \(\det(\mathbf{S}^{-1}(\mathbf{A}-t\mathbf{I}_n)\mathbf{S})=\det(\mathbf{S}^{-1})\det(\mathbf{A}-t\mathbf{I}_n)\det(\mathbf{S})=\det(\mathbf{A}-t\mathbf{I}_n)\). Thus \((\lambda-t)^g\) divides \(p_{\mathbf{A}}(t)\), so the root \(\lambda\) has multiplicity at least \(g\). Hence \(\operatorname{gm}(\lambda)\leq\operatorname{am}(\lambda)\). \(\square\)
The theorem gives an upper bound, not an equality in every case. A repeated root may have an eigenspace smaller than its algebraic multiplicity. For that reason, the number of times a root appears in the characteristic polynomial does not by itself tell us how many independent eigenvectors we can find.
Worked Example: A Repeated Root with Too Few Eigenvectors
Consider \(\mathbf{A}=\begin{bmatrix}2&1&0\\0&2&0\\0&0&5\end{bmatrix}\). Since \(\mathbf{A}-t\mathbf{I}_3\) is upper triangular, its determinant is the product of its diagonal entries:
Thus \(\operatorname{am}(2)=2\) and \(\operatorname{am}(5)=1\). For \(\lambda=2\), solve \((\mathbf{A}-2\mathbf{I}_3)\mathbf{x}=\mathbf{0}\), with \(\mathbf{x}=(x,y,z)^{\top}\):
So \(E_2=\operatorname{span}\{(1,0,0)^{\top}\}\) and \(\operatorname{gm}(2)=1\), not \(2\). For \(\lambda=5\), the equations from \((\mathbf{A}-5\mathbf{I}_3)\mathbf{x}=\mathbf{0}\) are \(-3x+y=0\), \(-3y=0\), and \(0=0\). They give \(x=y=0\), with \(z\) free, so \(E_5=\operatorname{span}\{(0,0,1)^{\top}\}\) and \(\operatorname{gm}(5)=1\).
Check: \(\mathbf{A}(1,0,0)^{\top}=(2,0,0)^{\top}=2(1,0,0)^{\top}\), and \(\mathbf{A}(0,0,1)^{\top}=(0,0,5)^{\top}=5(0,0,1)^{\top}\). These are eigenvectors, but there is only one independent eigenvector for eigenvalue \(2\).
Using Multiplicities to Test Diagonalizability
A real matrix is diagonalizable over \(\mathbb{R}\) if it has a basis of \(\mathbb{R}^n\) made of real eigenvectors. To use multiplicities as a diagonalizability test, we need to account for every eigenvalue. The theorem below assumes the characteristic polynomial splits into linear factors over \(\mathbb{R}\), so all its roots are real.
We use the fact that eigenspaces for distinct eigenvalues contribute independent vectors. For completeness, the fact and the criterion are proved here.
Proof. First suppose \(\lambda_1,\ldots,\lambda_k\) are distinct eigenvalues, and for each \(i\), let \(\mathbf{v}_i\in E_{\lambda_i}\). We show that if \(\mathbf{v}_1+\cdots+\mathbf{v}_k=\mathbf{0}\), then every \(\mathbf{v}_i=\mathbf{0}\). For \(k=1\), this follows directly from the equation. For \(k>1\), apply \(\mathbf{A}-\lambda_k\mathbf{I}_n\) to the equation. The term \(\mathbf{v}_k\) becomes zero, giving \(\sum_{i=1}^{k-1}(\lambda_i-\lambda_k)\mathbf{v}_i=\mathbf{0}\). By induction, each \((\lambda_i-\lambda_k)\mathbf{v}_i=\mathbf{0}\) for \(i<k\). Since the eigenvalues are distinct, \(\lambda_i-\lambda_k\ne0\), so \(\mathbf{v}_i=\mathbf{0}\) for \(i<k\). The original equation then gives \(\mathbf{v}_k=\mathbf{0}\). Thus, bases chosen from distinct eigenspaces have an independent union.
Now suppose first that \(\mathbf{A}\) is diagonalizable. It has a basis \(\mathbf{v}_1,\ldots,\mathbf{v}_n\) of eigenvectors, with corresponding eigenvalues \(d_1,\ldots,d_n\). In this basis, the matrix is diagonal with entries \(d_1,\ldots,d_n\), so \(p_{\mathbf{A}}(t)=\prod_{j=1}^n(d_j-t)\). Therefore \(\operatorname{am}(\lambda)\) is the number of basis vectors whose eigenvalue is \(\lambda\). Those vectors span all of \(E_{\lambda}\): if \(\mathbf{w}=\sum_j c_j\mathbf{v}_j\) satisfies \(\mathbf{A}\mathbf{w}=\lambda\mathbf{w}\), then \(\sum_j(d_j-\lambda)c_j\mathbf{v}_j=\mathbf{0}\). The basis vectors are independent, so \(c_j=0\) whenever \(d_j\ne\lambda\). Hence \(\operatorname{gm}(\lambda)=\operatorname{am}(\lambda)\).
Conversely, suppose \(\operatorname{gm}(\lambda)=\operatorname{am}(\lambda)\) for each eigenvalue. Because the characteristic polynomial splits, its distinct eigenvalues account for all \(n\) roots counted with multiplicity. Thus the sum of their algebraic multiplicities is \(n\), and the sum of their geometric multiplicities is also \(n\). Choose a basis for each eigenspace. By the independence result just proved, the union of these bases is independent; it contains \(n\) vectors in \(\mathbb{R}^n\), so it is a basis. This is an eigenvector basis, and therefore \(\mathbf{A}\) is diagonalizable. \(\square\)
The equality for each eigenvalue matters. Checking just one repeated eigenvalue, or noticing that one eigenspace is nontrivial, does not establish that the whole matrix has an eigenvector basis.
Worked Example: A Repeated Eigenvalue with Enough Eigenvectors
Let \(\mathbf{B}=\begin{bmatrix}2&0&1\\0&2&0\\0&0&5\end{bmatrix}\). Its characteristic polynomial is \((2-t)^2(5-t)\), so the algebraic multiplicities are \(2\) for \(2\) and \(1\) for \(5\). For eigenvalue \(2\), the equations from \((\mathbf{B}-2\mathbf{I}_3)\mathbf{x}=\mathbf{0}\) reduce to \(z=0\). Hence \(E_2=\operatorname{span}\{(1,0,0)^{\top},(0,1,0)^{\top}\}\), and \(\operatorname{gm}(2)=2\).
For eigenvalue \(5\), the equations are \(-3x+z=0\) and \(-3y=0\). Therefore \(E_5=\operatorname{span}\{(1,0,3)^{\top}\}\), and \(\operatorname{gm}(5)=1\). Both geometric multiplicities equal their algebraic multiplicities. The vectors \((1,0,0)^{\top}\), \((0,1,0)^{\top}\), and \((1,0,3)^{\top}\) are independent: a linear combination equal to zero has third coordinate \(3c=0\), then first and second coordinates force the other two coefficients to be zero.
Check: \(\mathbf{B}(1,0,0)^{\top}=2(1,0,0)^{\top}\), \(\mathbf{B}(0,1,0)^{\top}=2(0,1,0)^{\top}\), and \(\mathbf{B}(1,0,3)^{\top}=(5,0,15)^{\top}=5(1,0,3)^{\top}\). These three independent eigenvectors form a basis, as the criterion predicts.
A Parameter Can Change the Eigenspace Dimension
Worked Example: Same Algebraic Multiplicities, Different Outcome
For \(s\in\mathbb{R}\), consider \(\mathbf{C}_s=\begin{bmatrix}1&s&0\\0&1&0\\0&0&4\end{bmatrix}\). For every \(s\), the characteristic polynomial is \((1-t)^2(4-t)\). Thus the algebraic multiplicities stay fixed: \(\operatorname{am}(1)=2\) and \(\operatorname{am}(4)=1\).
For eigenvalue \(1\), the system \((\mathbf{C}_s-\mathbf{I}_3)\mathbf{x}=\mathbf{0}\) is \(sy=0\) and \(3z=0\). If \(s=0\), then \(z=0\) and both \(x\) and \(y\) are free, so \(E_1=\operatorname{span}\{(1,0,0)^{\top},(0,1,0)^{\top}\}\), with geometric multiplicity \(2\). If \(s\ne0\), then \(y=0\) and \(z=0\), so \(E_1=\operatorname{span}\{(1,0,0)^{\top}\}\), with geometric multiplicity \(1\). For eigenvalue \(4\), the equations give \(x=y=0\), so \(E_4=\operatorname{span}\{(0,0,1)^{\top}\}\), with geometric multiplicity \(1\), for every \(s\).
Check: \(\mathbf{C}_s(1,0,0)^{\top}=(1,0,0)^{\top}\) and \(\mathbf{C}_s(0,0,1)^{\top}=(0,0,4)^{\top}\) for all \(s\). When \(s=0\), \(\mathbf{C}_0(0,1,0)^{\top}=(0,1,0)^{\top}\) also, giving three independent eigenvectors. When \(s\ne0\), \(\mathbf{C}_s(0,1,0)^{\top}=(s,1,0)^{\top}\ne(0,1,0)^{\top}\), and the multiplicity criterion shows that \(\mathbf{C}_s\) is not diagonalizable.
Common Mistake: Treating a Root Count as an Eigenvector Count
For the first worked example, \(\operatorname{am}(2)=2\). A mistaken argument would be: “The eigenvalue \(2\) appears twice, so there are two independent eigenvectors for \(2\); therefore the matrix has enough eigenvectors to diagonalize.” But the equation for \(E_2\) requires \(y=0\) and \(z=0\), leaving only multiples of \((1,0,0)^{\top}\). For example, \((0,1,0)^{\top}\) is not an eigenvector with eigenvalue \(2\), because \(\mathbf{A}(0,1,0)^{\top}=(1,2,0)^{\top}\ne(0,2,0)^{\top}\). The correction is to compute the null space of \(\mathbf{A}-2\mathbf{I}_3\) and find its dimension; here that dimension is \(1\).
Keep the two checks separate: factor the characteristic polynomial to find algebraic multiplicities, and solve \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\) to find geometric multiplicities. Under the splitting hypothesis of the criterion, compare them eigenvalue by eigenvalue. Equality for every eigenvalue gives an eigenvector basis; a strict inequality for even one eigenvalue rules it out.
Check Your Understanding
For each problem, distinguish the polynomial root count from the eigenspace dimension.
- For \(\mathbf{A}=\begin{bmatrix}3&1\\0&3\end{bmatrix}\), find \(\operatorname{am}(3)\) and \(\operatorname{gm}(3)\).
- For \(\mathbf{B}=\begin{bmatrix}4&0\\0&4\end{bmatrix}\), find the algebraic and geometric multiplicities of \(4\). Is \(\mathbf{B}\) diagonalizable?
- Explain why \(\operatorname{gm}(\lambda)\leq\operatorname{am}(\lambda)\) does not imply equality.
- State the splitting condition needed to apply the real multiplicity criterion as written.
- A real \(3\times3\) matrix has two eigenvalues, with algebraic multiplicities \(2\) and \(1\). Their geometric multiplicities are \(1\) and \(1\). Is the matrix diagonalizable over \(\mathbb{R}\)? Explain.
