Two Ways to Count an Eigenvalue
An eigenvalue can be counted in two different ways. Its algebraic multiplicity records how many times it occurs as a root of the characteristic polynomial. Its geometric multiplicity records how many independent directions belong to its eigenspace. For a general matrix, these counts need not agree. For a real symmetric matrix, orthogonal diagonalization shows exactly why they do: each occurrence of an eigenvalue on the diagonal contributes one independent eigenvector direction and one factor to the characteristic polynomial.
Throughout this tutorial, \(\mathbf{A}\) is a real \(n\times n\) matrix. An eigenvalue \(\lambda\) of \(\mathbf{A}\) is a real number for which \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\) for some nonzero \(\mathbf{x}\in\mathbb{R}^n\). Its eigenspace is \(E_{\lambda}=\{\mathbf{x}\in\mathbb{R}^n:\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\}\), which includes the zero vector as well as all eigenvectors with eigenvalue \(\lambda\). The characteristic polynomial follows the course convention \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I})\).
The result in this tutorial applies to real symmetric matrices: matrices satisfying \(\mathbf{A}^{\top}=\mathbf{A}\). We use the Spectral Theorem, established earlier in the course: every real symmetric \(n\times n\) matrix has an orthogonal diagonalization \(\mathbf{A}=\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top}\), where \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{Q}\mathbf{Q}^{\top}=\mathbf{I}_n\) and \(\mathbf{D}=\operatorname{diag}(d_1,\dots,d_n)\) is diagonal. The columns of \(\mathbf{Q}\) are an orthonormal eigenbasis of \(\mathbf{A}\), and the corresponding entries \(d_j\) are their eigenvalues.
What the Diagonalization Says About an Eigenspace
Orthogonality makes the change of coordinates particularly direct. If \(\mathbf{x}\) is a vector in the original coordinates, then \(\mathbf{y}=\mathbf{Q}^{\top}\mathbf{x}\) is its coordinate vector in the orthonormal eigenbasis, and \(\mathbf{x}=\mathbf{Q}\mathbf{y}\). Because \(\mathbf{Q}^{\top}\) is the inverse of \(\mathbf{Q}\), this correspondence is reversible; it cannot lose or add independent directions.
Proof. Let \(\mathbf{x}\in\mathbb{R}^n\) and set \(\mathbf{y}=\mathbf{Q}^{\top}\mathbf{x}\), so \(\mathbf{x}=\mathbf{Q}\mathbf{y}\). Since \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\), the equation \(\mathbf{A}\mathbf{x}=\mu\mathbf{x}\) is equivalent, after multiplying on the left by \(\mathbf{Q}^{\top}\), to \(\mathbf{D}\mathbf{y}=\mu\mathbf{y}\). In coordinate \(j\), this is \(d_jy_j=\mu y_j\), or \((d_j-\mu)y_j=0\). If \(d_j\ne\mu\), this forces \(y_j=0\); if \(d_j=\mu\), it places no restriction on \(y_j\). Thus \(\mathbf{y}\) is any linear combination of the standard basis vectors \(\mathbf{e}_j\) with \(j\in J_{\mu}\), and \(\mathbf{x}=\mathbf{Q}\mathbf{y}\) gives the stated description of \(E_{\mu}\). The vectors \(\mathbf{Q}\mathbf{e}_j\) for \(j\in J_{\mu}\) are linearly independent because \(\mathbf{Q}\) is invertible, and they span \(E_{\mu}\) by that description. They are therefore a basis, including the case \(J_{\mu}\) is empty, when the eigenspace is \(\{\mathbf{0}\}\) and has dimension zero. \(\square\)
The proposition describes all eigenspace vectors, not only a list of eigenvectors to select from. If \(\mu\) appears several times on the diagonal, any linear combination of the corresponding basis vectors is also in \(E_{\mu}\). The individual vectors \(\mathbf{Q}\mathbf{e}_j\) form a basis of that space; they are not the only eigenvectors in it.
Why the Characteristic Polynomial Counts the Same Entries
The characteristic polynomial is unchanged by the orthogonal change of coordinates. In fact, \(\mathbf{A}-\lambda\mathbf{I}_n=\mathbf{Q}(\mathbf{D}-\lambda\mathbf{I}_n)\mathbf{Q}^{\top}\). Using the determinant product rule and \(\det(\mathbf{Q}^{\top})\det(\mathbf{Q})=\det(\mathbf{Q}^{\top}\mathbf{Q})=\det(\mathbf{I}_n)=1\), we obtain \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{D}-\lambda\mathbf{I}_n)\). The determinant of a diagonal matrix is the product of its diagonal entries, so the characteristic polynomial factors into one linear factor for each diagonal entry.
Proof. By the Spectral Theorem, write \(\mathbf{A}=\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top}\), where \(\mathbf{D}=\operatorname{diag}(d_1,\dots,d_n)\). The determinant calculation above gives \(p_{\mathbf{A}}(\lambda)=\prod_{j=1}^{n}(d_j-\lambda)\). The multiplicity of \(\mu\) as a root of this product is exactly the number of indices \(j\) for which \(d_j=\mu\): each such index contributes one factor \(d_j-\lambda=\mu-\lambda\), and all other factors have nonzero value at \(\lambda=\mu\). Hence \(\operatorname{am}(\mu)=|J_{\mu}|\). By the proposition, the vectors \(\mathbf{Q}\mathbf{e}_j\) with \(j\in J_{\mu}\) form a basis of \(E_{\mu}\), so \(\operatorname{gm}(\mu)=\dim E_{\mu}=|J_{\mu}|\). The two multiplicities are equal. \(\square\)
This argument also accounts for every eigenvalue: the proposition shows that an eigenvalue must equal some diagonal entry \(d_j\), since a nonzero vector in its eigenspace requires at least one coordinate \(y_j\ne0\) with \(d_j=\mu\). Conversely, if \(d_j=\mu\), then \(\mathbf{Q}\mathbf{e}_j\ne\mathbf{0}\) is an eigenvector for \(\mu\). Thus the diagonal entries list all eigenvalues, with repetitions that record both multiplicities.
Worked Consequences of the Theorem
Worked Example: Two Distinct Eigenvalues
Consider the symmetric matrix \(\mathbf{A}=\begin{bmatrix}3&1\\1&3\end{bmatrix}\). Find its eigenvalues and compare algebraic and geometric multiplicities.
First compute the characteristic polynomial:
The eigenvalues are \(2\) and \(4\), each with algebraic multiplicity \(1\). To find their eigenspaces, for \(\lambda=2\) solve \((\mathbf{A}-2\mathbf{I}_2)\mathbf{x}=\mathbf{0}\): this gives \(x_1+x_2=0\), so \(E_2=\operatorname{span}\{(1,-1)^{\top}\}\). For \(\lambda=4\), the equation gives \(-x_1+x_2=0\), so \(E_4=\operatorname{span}\{(1,1)^{\top}\}\). Each eigenspace has dimension \(1\).
Check: \(\mathbf{A}(1,-1)^{\top}=(2,-2)^{\top}=2(1,-1)^{\top}\), and \(\mathbf{A}(1,1)^{\top}=(4,4)^{\top}=4(1,1)^{\top}\). The vectors are nonzero and independent, since \((1,-1)^{\top}\) is not a scalar multiple of \((1,1)^{\top}\). The computed eigenspaces and polynomial factors agree: both eigenvalues have algebraic and geometric multiplicity \(1\).
Worked Example: A Repeated Eigenvalue in a Rotated Eigenspace
Let \(\mathbf{Q}=\begin{bmatrix}3/5&0&-4/5\\0&1&0\\4/5&0&3/5\end{bmatrix}\) and \(\mathbf{D}=\operatorname{diag}(2,2,7)\). The columns of \(\mathbf{Q}\) have length \(1\), and their pairwise dot products are \(0\); hence \(\mathbf{Q}\) is orthogonal. Form \(\mathbf{A}=\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top}\).
Writing the columns of \(\mathbf{Q}\) as \(\mathbf{q}_1,\mathbf{q}_2,\mathbf{q}_3\), the product is \(2\mathbf{q}_1\mathbf{q}_1^{\top}+2\mathbf{q}_2\mathbf{q}_2^{\top}+7\mathbf{q}_3\mathbf{q}_3^{\top}\). Its entries give
Orthogonal diagonalization gives \(p_{\mathbf{A}}(\lambda)=(2-\lambda)^2(7-\lambda)\). Therefore \(2\) has algebraic multiplicity \(2\), while \(7\) has algebraic multiplicity \(1\). The eigenspace description gives \(E_2=\operatorname{span}\{(3/5,0,4/5)^{\top},(0,1,0)^{\top}\}\), of dimension \(2\), and \(E_7=\operatorname{span}\{(-4/5,0,3/5)^{\top}\}\), of dimension \(1\).
Check: Multiplying the displayed matrix by the first basis vector for \(E_2\) gives \((6/5,0,8/5)^{\top}=2(3/5,0,4/5)^{\top}\); multiplying by the second gives \((0,2,0)^{\top}=2(0,1,0)^{\top}\). For the vector in \(E_7\), the product is \((-28/5,0,21/5)^{\top}=7(-4/5,0,3/5)^{\top}\). Moreover, the sum of the first two vectors in the basis of \(E_2\) is another eigenvector, \((3/5,1,4/5)^{\top}\), since its image is \((6/5,2,8/5)^{\top}\), twice that vector. Thus the basis vectors span the eigenspace; they are not its only eigenvectors. The dimensions \(2\) and \(1\) match the polynomial multiplicities.
Why Symmetry Is Essential
The equality is not a rule for all matrices. Orthogonal diagonalization is what makes the two counts coincide: each repeated diagonal entry appears once per basis direction and once per characteristic-polynomial factor. Without symmetry, a repeated polynomial root does not necessarily supply enough independent eigenvectors.
Worked Example: A Repeated Root Without Enough Eigenvectors
Consider the nonsymmetric matrix \(\mathbf{B}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\). Its characteristic polynomial is \(p_{\mathbf{B}}(\lambda)=\det\begin{bmatrix}1-\lambda&1\\0&1-\lambda\end{bmatrix}=(1-\lambda)^2\). Thus its only eigenvalue is \(1\), with algebraic multiplicity \(2\).
To find its eigenspace, solve \((\mathbf{B}-\mathbf{I}_2)\mathbf{x}=\mathbf{0}\). The equation is \(x_2=0\), so \(E_1=\operatorname{span}\{(1,0)^{\top}\}\), which has dimension \(1\). Directly, \(\mathbf{B}(1,0)^{\top}=(1,0)^{\top}\). Therefore \(\operatorname{gm}(1)=1\), not \(2\). This does not contradict the theorem: \(\mathbf{B}\) is not symmetric, since its off-diagonal entries differ.
Common Pitfalls and a Reliable Check
Pitfall: treating algebraic multiplicity as a list of individual eigenvectors. If an eigenvalue appears twice, that does not mean there are exactly two eigenvectors. There are infinitely many nonzero vectors in a two-dimensional eigenspace, because every nonzero linear combination of two independent eigenvectors in that space is also an eigenvector. In the rotated example, the two displayed vectors form a basis for \(E_2\), and their nonzero combinations give further eigenvectors.
Pitfall: applying the equality without its hypothesis. The calculation for \(\mathbf{B}\) would be misread if one concluded that a root of multiplicity \(2\) must have a two-dimensional eigenspace for every matrix. The theorem requires a real symmetric matrix. For a symmetric matrix, first use an orthogonal diagonalization; then count the occurrences of the eigenvalue on the diagonal. The polynomial count and the eigenspace dimension are both that same count.
For a real symmetric \(n\times n\) matrix, the multiplicities of all distinct eigenvalues add to \(n\): the diagonal matrix in an orthogonal diagonalization has \(n\) entries, and each entry contributes once to exactly one eigenvalue's algebraic multiplicity and one basis vector to exactly one eigenspace. This is a useful consistency check, but it depends on counting repeated entries rather than only distinct eigenvalues.
Check Your Understanding
For each question, distinguish the polynomial root count from the dimension of the eigenspace.
- If a real symmetric matrix is orthogonally diagonalized with diagonal entries \(3,3,-2,5\), state the algebraic and geometric multiplicities of each eigenvalue.
- For a symmetric matrix with \(\mathbf{A}=\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top}\), explain which vectors form a basis of \(E_{\mu}\) when \(\mu\) occurs in diagonal positions \(2\) and \(4\).
- Why does \(\mathbf{Q}\) being orthogonal ensure that changing from \(\mathbf{x}\) to \(\mathbf{y}=\mathbf{Q}^{\top}\mathbf{x}\) preserves the dimension of an eigenspace?
- A characteristic polynomial has a root \(\mu\) of multiplicity \(3\) for a real symmetric matrix. What is \(\dim E_{\mu}\), and which theorem justifies the conclusion?
- For the matrix \(\mathbf{B}\) in the last example, identify its algebraic and geometric multiplicities for eigenvalue \(1\), and state the hypothesis of the theorem that fails.
