Why a Repeated Eigenvalue Needs Care
A repeated root of the characteristic polynomial tells us how many times an eigenvalue occurs algebraically. It does not, by itself, tell us how many independent eigenvectors belong to that eigenvalue. To diagonalize a matrix, we need a basis of eigenvectors, so a repeated eigenvalue must contribute enough independent vectors to that basis.
Throughout, matrices and vectors are real. For \(\mathbf{A}\in\mathbb{R}^{n\times n}\), 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 is \(E_{\lambda}=\operatorname{Nul}(\mathbf{A}-\lambda\mathbf{I}_n)\). The characteristic polynomial is \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_n)\).
To find \(E_{\lambda}\), solve \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\). Every solution is an eigenvector when it is nonzero. In particular, the zero vector belongs to the eigenspace as a subspace, but it is never an eigenvector. The free variables in the RREF of \(\mathbf{A}-\lambda\mathbf{I}_n\) give the number of independent directions in the eigenspace.
Proof. Row operations preserve the solution set of a homogeneous system, so we may solve using its RREF. Each pivot variable is determined by the free variables. Assigning one free variable the value \(1\) and all other free variables \(0\) produces one solution for each free variable. These solutions are independent, because each has value \(1\) in its own free-variable position and \(0\) in the other such positions. Every solution is a linear combination of these solutions: use its free-variable values as the coefficients, since the pivot variables are determined by those values. Thus these solutions form a basis of the solution space, and its dimension is the number of free variables. \(\square\)
From the Eigenspace to a Diagonalization
If \(\operatorname{am}(\lambda)=r\) and \(\operatorname{gm}(\lambda)=r\), the eigenspace has \(r\) independent directions. Choose \(r\) independent eigenvectors from it; do not list the same direction more than once. Those vectors supply the \(r\) columns associated with \(\lambda\) in the diagonal matrix. To diagonalize the whole matrix, the remaining columns must also be eigenvectors, and the complete list must be a basis of \(\mathbb{R}^n\).
Proof. Because the vectors form a basis, the matrix \(\mathbf{P}\) with those columns is invertible. The \(j\)th column of \(\mathbf{A}\mathbf{P}\) is \(\mathbf{A}\mathbf{v}_j=\lambda_j\mathbf{v}_j\), which is the \(j\)th column of \(\mathbf{P}\mathbf{D}\). Hence \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\). Multiplying on the right by \(\mathbf{P}^{-1}\) gives \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\). \(\square\)
This result explains the construction: the columns of \(\mathbf{P}\) are eigenvectors, while the corresponding eigenvalues, in the same order, form the diagonal entries of \(\mathbf{D}\). Repeating an eigenvalue on the diagonal requires a matching number of independent columns from its eigenspace.
A Three-by-Three Matrix with a Full Repeated Eigenspace
Worked Example: Two Independent Eigenvectors for \(\lambda=2\)
Diagonalize the matrix
The matrix is upper triangular, so its characteristic polynomial is the product of its diagonal entries after subtracting \(\lambda\): \(p_{\mathbf{A}}(\lambda)=(2-\lambda)^2(-1-\lambda)\). Thus \(\lambda=2\) has algebraic multiplicity \(2\), and \(\lambda=-1\) has algebraic multiplicity \(1\).
For \(\lambda=2\), row-reduce \(\mathbf{A}-2\mathbf{I}_3\):
The first step subtracts row 1 from rows 2 and 3; the second scales row 1 by \(-\tfrac13\). The equation is \(x_3=0\), with \(x_1,x_2\) free. Therefore \(E_2=\operatorname{span}\{(1,0,0),(0,1,0)\}\). Its dimension is \(2\), matching the algebraic multiplicity.
For \(\lambda=-1\), solving \((\mathbf{A}+\mathbf{I}_3)\mathbf{x}=\mathbf{0}\) gives the equations \(3x_1-3x_3=0\), \(3x_2-3x_3=0\), and no further condition. Taking \(x_3=1\) gives the eigenvector \((1,1,1)\).
Choose these three eigenvectors as the columns of \(\mathbf{P}\), in the same order as the diagonal entries of \(\mathbf{D}\):
The matrix \(\mathbf{P}\) is upper triangular with diagonal entries \(1,1,1\), so \(\det\mathbf{P}=1\) and it is invertible. Directly checking the columns gives
For example, the third column check is \(\mathbf{A}(1,1,1)^{\top}=(-1,-1,-1)^{\top}\). Thus \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\). The repeated eigenvalue appears twice in \(\mathbf{D}\), paired with the two independent vectors from \(E_2\).
A Full Eigenspace in Four Dimensions
Worked Example: A Repeated Eigenvalue in a Four-by-Four Matrix
Consider
Its characteristic polynomial is \((4-\lambda)^2(1-\lambda)(-2-\lambda)\), so \(\lambda=4\) has algebraic multiplicity \(2\). To find its eigenspace, reduce \(\mathbf{A}-4\mathbf{I}_4\):
For the second matrix, row 3 was replaced by row 3 plus three times row 1, and row 4 was scaled by \(-\tfrac16\). The pivot columns are 3 and 4, so \(x_1,x_2\) are free and \(x_3=x_4=0\). A basis of \(E_4\) is \(\mathbf{e}_1,\mathbf{e}_2\), giving geometric multiplicity \(2\).
The other eigenvalues have eigenvectors \((-1,0,3,0)^{\top}\) for \(\lambda=1\) and \(\mathbf{e}_4\) for \(\lambda=-2\). Indeed, multiplying by \(\mathbf{A}\) gives \((-1,0,3,0)^{\top}\) and \(-2\mathbf{e}_4\), respectively. Set
The determinant of \(\mathbf{P}\) is \(1\cdot1\cdot3\cdot1=3\), so \(\mathbf{P}\) is invertible. Multiplication column by column verifies \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\): the columns of the product are \((4,0,0,0)^{\top}\), \((0,4,0,0)^{\top}\), \((-1,0,3,0)^{\top}\), and \((0,0,0,-2)^{\top}\). Therefore \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\).
Choosing a Basis from a Non-coordinate Eigenspace
Worked Example: Finding Two Independent Vectors in a Plane
Let
The characteristic polynomial is \((-1-\lambda)(2-\lambda)^2\), so \(\lambda=2\) is repeated. The eigenspace equation comes from
Thus \(x_1-x_2+x_3=0\). Let \(x_2=s\) and \(x_3=t\); then \(x_1=s-t\), so every vector in \(E_2\) has the form \((s-t,s,t)^{\top}=s(1,1,0)^{\top}+t(0,1,1)^{\top}\). The vectors \((1,1,0)^{\top}\) and \((0,1,1)^{\top}\) are independent: if \(a(1,1,0)+b(0,1,1)=\mathbf{0}\), the first and third coordinates give \(a=0\) and \(b=0\). They therefore form a basis of the two-dimensional eigenspace.
For \(\lambda=-1\), the vector \(\mathbf{e}_1\) is an eigenvector because \(\mathbf{B}\mathbf{e}_1=(-1,0,0)^{\top}=-\mathbf{e}_1\). Put the chosen eigenvectors into the columns of \(\mathbf{P}\):
Expansion along the first row gives \(\det\mathbf{P}=1\). The column products are \(\mathbf{B}(1,1,0)^{\top}=(2,2,0)^{\top}\), \(\mathbf{B}(0,1,1)^{\top}=(0,2,2)^{\top}\), and \(\mathbf{B}\mathbf{e}_1=(-1,0,0)^{\top}\); hence \(\mathbf{B}\mathbf{P}=\mathbf{P}\mathbf{D}\). Since \(\mathbf{P}\) is invertible, \(\mathbf{B}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\).
Common Pitfall: Counting a Direction Twice
A repeated eigenvalue does not mean that any list of the same length from its eigenspace will work. For the first example, a mistaken choice for the two columns associated with \(\lambda=2\) would be \(\mathbf{e}_1\) and \(2\mathbf{e}_1\). They are both eigenvectors, but they are dependent: \(2\mathbf{e}_1\) is a scalar multiple of \(\mathbf{e}_1\). Using them as two columns makes the proposed eigenvector matrix singular, not invertible.
The correction is to choose two independent directions in \(E_2\), such as \(\mathbf{e}_1,\mathbf{e}_2\). Their independence can be checked directly: \(a\mathbf{e}_1+b\mathbf{e}_2=\mathbf{0}\) forces \(a=b=0\). More generally, when \(\operatorname{am}(\lambda)=r\), find the eigenspace by row-reducing \(\mathbf{A}-\lambda\mathbf{I}_n\), then choose \(r\) independent vectors from its solution space if it has dimension \(r\). Only after including eigenvectors for the other eigenvalues and confirming that all \(n\) columns form a basis may you conclude that the full matrix is diagonalizable.
This distinction is why both multiplicities matter. Algebraic multiplicity tells how often \(\lambda\) is a root; geometric multiplicity counts the independent eigenvector directions available for the diagonalization. Matching the repeated root with a basis of its eigenspace gives the correct repeated-eigenvalue columns in \(\mathbf{D}\) and \(\mathbf{P}\).
Check Your Understanding
For each question, use the eigenspace equation and check proposed eigenvectors by multiplication.
- For the first worked example, what are the free variables in the RREF of \(\mathbf{A}-2\mathbf{I}_3\), and what is a basis of \(E_2\)?
- In the four-dimensional example, why does the RREF of \(\mathbf{A}-4\mathbf{I}_4\) show that the geometric multiplicity of \(4\) is \(2\)?
- For \(\mathbf{B}\) in the third example, verify directly that \((1,1,0)^{\top}\) and \((0,1,1)^{\top}\) are eigenvectors for \(\lambda=2\).
- Explain why two nonzero eigenvectors for the same eigenvalue do not necessarily provide two independent columns of \(\mathbf{P}\).
- State what must be true of the full list of columns of \(\mathbf{P}\) before \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\) is valid.
