The Repeated-Root Case in Dimension Three
When a real \(3\times3\) matrix has three distinct eigenvalues, each eigenvalue supplies an eigenvector, and those eigenvectors form a basis. A repeated eigenvalue changes the question: it counts twice in the characteristic polynomial, but its eigenspace may or may not contain two independent eigenvectors. The decisive computation is to find the dimensions of the eigenspaces, not merely to count the roots.
Let \(\mathbf{A}\in\mathbb{R}^{3\times3}\). A real number \(\lambda\) is an eigenvalue if there is a nonzero vector \(\mathbf{v}\) such that \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). Its eigenspace is \(E_\lambda=\operatorname{Nul}(\mathbf{A}-\lambda\mathbf{I}_3)\), including the zero vector. The characteristic polynomial is \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_3)\). The algebraic multiplicity \(\operatorname{am}(\lambda)\) is the multiplicity of \(\lambda\) as a root of that polynomial; the geometric multiplicity \(\operatorname{gm}(\lambda)\) is \(\dim E_\lambda\).
This criterion uses the standard multiplicity result established in the earlier material on algebraic and geometric multiplicities: for each eigenvalue, \(1\leq\operatorname{gm}(\lambda)\leq\operatorname{am}(\lambda)\), and a \(3\times3\) matrix is diagonalizable exactly when the sum of the geometric multiplicities of its distinct eigenvalues is \(3\). Since \(\operatorname{am}(\beta)=1\), we have \(\operatorname{gm}(\beta)=1\). Thus the only unresolved question in this pattern is whether \(\dim E_\alpha\) is \(2\) or \(1\).
Proof. If \(\dim E_\alpha=2\), choose a basis of \(E_\alpha\) and a nonzero vector in \(E_\beta\). The multiplicity result gives \(\dim E_\beta=1\), so the sum of the eigenspace dimensions is \(2+1=3\). The multiplicity criterion therefore gives a basis of eigenvectors, and \(\mathbf{A}\) is diagonalizable. Conversely, if \(\mathbf{A}\) is diagonalizable, it has a basis of three eigenvectors. Each basis vector belongs to \(E_\alpha\) or \(E_\beta\). Since \(\dim E_\beta=1\), at most one independent basis vector can belong to \(E_\beta\); the other two must belong to \(E_\alpha\). Hence \(\dim E_\alpha\geq2\), and the multiplicity bound \(\dim E_\alpha\leq\operatorname{am}(\alpha)=2\) forces \(\dim E_\alpha=2\). \(\square\)
A Repeated Eigenvalue With Enough Eigenvectors
Worked Example: A Two-Dimensional Eigenspace
Determine whether \(\mathbf{A}=\begin{bmatrix}3&1&1\\1&3&1\\1&1&3\end{bmatrix}\) is diagonalizable over \(\mathbb{R}\), and find an eigenbasis if it is.
First compute its characteristic polynomial. Set \(a=3-\lambda\). Expanding the determinant along the first row gives
Thus \(2\) is repeated and \(5\) is simple. To find \(E_2\), solve \((\mathbf{A}-2\mathbf{I}_3)\mathbf{x}=\mathbf{0}\):
The equation is \(x_1+x_2+x_3=0\). Taking \(x_1\) and \(x_2\) as free variables gives \(x_3=-x_1-x_2\), so a basis is \(\mathbf{v}_1=(1,-1,0)^{\top}\), \(\mathbf{v}_2=(1,0,-1)^{\top}\).
For \(E_5\), row-reduce \(\mathbf{A}-5\mathbf{I}_3\). Swap \(R_1\) and \(R_2\), then use \(R_2\to R_2+2R_1\), \(R_3\to R_3-R_1\), \(R_2\to-\tfrac13R_2\), \(R_3\to R_3+3R_2\), and \(R_1\to R_1+2R_2\):
Thus \(x_1=x_3\) and \(x_2=x_3\), so \(E_5=\operatorname{span}\{(1,1,1)^{\top}\}\). There are two independent eigenvectors for \(2\) and one for \(5\), giving three eigenvectors in total.
Check: Direct multiplication gives \(\mathbf{A}\mathbf{v}_1=2\mathbf{v}_1\), \(\mathbf{A}\mathbf{v}_2=2\mathbf{v}_2\), and \(\mathbf{A}(1,1,1)^{\top}=(5,5,5)^{\top}\). The matrix \(\mathbf{P}\) with these eigenvectors as columns has determinant \(3\), so the columns are independent. With \(\mathbf{D}=\operatorname{diag}(2,2,5)\), the three products give \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\), and therefore \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\).
Why Distinct Eigenspaces Contribute Independently
In the examples, eigenvectors from the two different eigenvalues cannot duplicate one another’s contribution. The following short argument explains why. It also justifies combining bases from the two eigenspaces when checking the total number of independent eigenvectors.
Proof. Suppose \(\mathbf{v}\in E_\alpha\cap E_\beta\). By the definition of eigenspace, \(\mathbf{A}\mathbf{v}=\alpha\mathbf{v}\) and \(\mathbf{A}\mathbf{v}=\beta\mathbf{v}\). Subtracting gives \((\alpha-\beta)\mathbf{v}=\mathbf{0}\). Since \(\alpha-\beta\ne0\), division by this scalar gives \(\mathbf{v}=\mathbf{0}\). Thus the intersection contains only the zero vector. If nonzero vectors \(\mathbf{u}\in E_\alpha\) and \(\mathbf{v}\in E_\beta\) were dependent, one would be a scalar multiple of the other; that nonzero vector would then belong to both eigenspaces, contradicting the intersection result. \(\square\)
For the repeated-root pattern, this independence statement is consistent with the multiplicity criterion: a basis of the two-dimensional repeated eigenspace and a nonzero vector from the simple eigenspace together give three independent eigenvectors. The dimension calculation still matters. Distinct eigenvalues keep their eigenspaces separate, but they do not make a one-dimensional repeated eigenspace larger.
A Repeated Eigenvalue With Too Few Eigenvectors
Worked Example: The Repeated Root Is Not Enough
Test \(\mathbf{B}=\begin{bmatrix}2&1&0\\0&2&0\\0&0&5\end{bmatrix}\) for diagonalizability over \(\mathbb{R}\).
This matrix is upper triangular, so its characteristic polynomial is the product of its diagonal entries after subtracting \(\lambda\mathbf{I}_3\):
The eigenvalues and algebraic multiplicities match the previous pattern. For \(\lambda=2\), solve \((\mathbf{B}-2\mathbf{I}_3)\mathbf{x}=\mathbf{0}\):
Only \(x_1\) is free, so \(E_2=\operatorname{span}\{(1,0,0)^{\top}\}\) and \(\dim E_2=1\), not \(2\). For \(\lambda=5\), the equations are \(-3x_1+x_2=0\), \(-3x_2=0\), and \(0=0\). The second equation gives \(x_2=0\), and then the first gives \(x_1=0\); \(x_3\) is free. Hence \(E_5=\operatorname{span}\{(0,0,1)^{\top}\}\).
Check: \(\mathbf{B}(1,0,0)^{\top}=(2,0,0)^{\top}\) and \(\mathbf{B}(0,0,1)^{\top}=(0,0,5)^{\top}\), as required. These are the only two one-dimensional eigenspaces, so their eigenvectors cannot make a basis of \(\mathbb{R}^3\). The matrix is not diagonalizable.
Another Full Eigenspace Computation
Worked Example: Reading the Eigenspaces Directly
Decide whether \(\mathbf{C}=\begin{bmatrix}2&0&0\\0&2&3\\0&0&5\end{bmatrix}\) is diagonalizable, and verify a diagonalization.
Its characteristic polynomial is \((2-\lambda)^2(5-\lambda)\), so the repeated eigenvalue is \(2\). For \(E_2\), the system is \(3x_3=0\), so \(x_3=0\), while \(x_1\) and \(x_2\) are free. A basis is \((1,0,0)^{\top},(0,1,0)^{\top}\).
For \(E_5\), the equations from \(\mathbf{C}-5\mathbf{I}_3\) are \(-3x_1=0\), \(-3x_2+3x_3=0\), and \(0=0\). Therefore \(x_1=0\), \(x_2=x_3\), and a basis is \((0,1,1)^{\top}\). The eigenspace dimensions are \(2\) and \(1\), so the matrix is diagonalizable.
Check: Set \(\mathbf{P}=\begin{bmatrix}1&0&0\\0&1&1\\0&0&1\end{bmatrix}\) and \(\mathbf{D}=\operatorname{diag}(2,2,5)\). The determinant of \(\mathbf{P}\) is \(1\), and multiplying by columns gives
Thus \(\mathbf{C}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\), with each column of \(\mathbf{P}\) verified as an eigenvector in the matching position of \(\mathbf{D}\).
Common Mistake: Counting a Root Twice as a Vector
A common incorrect argument is: “The characteristic polynomial has a repeated root, so that eigenvalue gives two eigenvectors.” This confuses algebraic multiplicity with geometric multiplicity. In the example with \(\mathbf{B}\), the root \(2\) occurs twice, but the equations \(x_2=0\) and \(x_3=0\) leave only one free variable. Its eigenspace is one-dimensional.
The correction is to solve \((\mathbf{A}-\lambda\mathbf{I}_3)\mathbf{x}=\mathbf{0}\) for each eigenvalue and count the free variables, or otherwise find a basis of each null space. In this setting the decision is exact: with characteristic polynomial \((\alpha-\lambda)^2(\beta-\lambda)\), diagonalizability holds if and only if the repeated eigenvalue has a two-dimensional eigenspace. A repeated root with only one eigenvector gives just two independent eigenvectors in total, so it cannot provide an eigenbasis for \(\mathbb{R}^3\).
Check Your Understanding
For each question, distinguish the multiplicity of a root from the dimension of its eigenspace.
- A real \(3\times3\) matrix has characteristic polynomial \((3-\lambda)^2(1-\lambda)\). What dimension must \(E_3\) have for the matrix to be diagonalizable?
- For \(\mathbf{A}=\begin{bmatrix}4&0&0\\0&4&1\\0&0&-2\end{bmatrix}\), solve for \(E_4\). Is the matrix diagonalizable?
- Explain why an eigenvalue of algebraic multiplicity \(1\) must have a one-dimensional eigenspace.
- For the matrix \(\mathbf{B}\) in the second worked example, verify directly that \((1,0,0)^{\top}\) and \((0,0,1)^{\top}\) are eigenvectors, then explain why they are not an eigenbasis.
- State the test for diagonalizability when a real \(3\times3\) matrix has two distinct eigenvalues and one has algebraic multiplicity \(2\).
