Why Triangular Matrices Are Easier to Analyze
A triangular matrix has zeros on one side of its diagonal. This structure makes its eigenvalues especially easy to identify: the diagonal entries are exactly the eigenvalues, including any repetitions. Finding eigenvectors still requires solving equations, however. In particular, repeated diagonal entries do not by themselves tell us whether the matrix is diagonalizable; we must check whether there are enough linearly independent eigenvectors.
Throughout, take \(\mathbf{A}\in\mathbb{R}^{n\times n}\). An upper triangular matrix has \(a_{ij}=0\) when \(i>j\), and a lower triangular matrix has \(a_{ij}=0\) when \(i<j\). An eigenvalue \(\lambda\) has a nonzero eigenvector \(\mathbf{v}\) when \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). The eigenspace is \(E_{\lambda}=\{\mathbf{v}:(\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{v}=\mathbf{0}\}\); it includes the zero vector as well as all eigenvectors for \(\lambda\).
Proof. The matrix \(\mathbf{A}-\lambda\mathbf{I}_n\) is triangular with diagonal entries \(a_{ii}-\lambda\). In the determinant expansion, a nonzero term for an upper triangular matrix would require the selected column \(\sigma(i)\) in row \(i\) to satisfy \(\sigma(i)\ge i\) for every \(i\). Since \(\sum_i\sigma(i)=\sum_i i\), all these inequalities must be equalities, so \(\sigma(i)=i\) for every \(i\). Thus only the diagonal product can contribute. For a lower triangular matrix the inequalities are reversed, \(\sigma(i)\le i\), and the same sum argument again forces \(\sigma(i)=i\). The determinant is therefore the displayed product in either case. A real number \(\lambda\) is an eigenvalue exactly when \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{v}=\mathbf{0}\) has a nonzero solution, which, by the determinant criterion for a square homogeneous system, is equivalent to \(\det(\mathbf{A}-\lambda\mathbf{I}_n)=0\). The roots and their multiplicities now follow from the product. \(\square\)
This result identifies the eigenvalues but does not provide their eigenvectors. For each distinct diagonal value, substitute that value into \(\mathbf{A}-\lambda\mathbf{I}_n\) and solve the resulting homogeneous system. A repeated diagonal value contributes the same number of factors to the characteristic polynomial as its number of occurrences; that number is not automatically the number of independent eigenvectors available.
From Eigenspaces to Diagonalization
A matrix is diagonalizable over \(\mathbb{R}\) if it is similar to a diagonal matrix: there are an invertible matrix \(\mathbf{P}\) and a diagonal matrix \(\mathbf{D}\) such that \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\). Equivalently, \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\). When the columns of \(\mathbf{P}\) are eigenvectors, this equation says that each column is multiplied by its corresponding diagonal entry of \(\mathbf{D}\).
Proof. If \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\), then \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\). Each column of \(\mathbf{P}\) is therefore an eigenvector, and its columns form a basis because \(\mathbf{P}\) is invertible. Conversely, if the columns of \(\mathbf{P}\) form an eigenvector basis, let the corresponding eigenvalues, in the same order, be the diagonal entries of \(\mathbf{D}\). Then \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\), and multiplying on the right by \(\mathbf{P}^{-1}\) gives the required diagonalization.
For the multiplicity statement, first note that eigenspaces for distinct eigenvalues have independent contributions. Indeed, suppose \(\mathbf{v}_1+\cdots+\mathbf{v}_s=\mathbf{0}\), where \(\mathbf{v}_j\in E_{\lambda_j}\). For \(s>1\), apply \(\mathbf{A}-\lambda_s\mathbf{I}_n\) to obtain \(\sum_{j=1}^{s-1}(\lambda_j-\lambda_s)\mathbf{v}_j=\mathbf{0}\). Induction on \(s\), starting with the immediate one-eigenspace case, shows that each \(\mathbf{v}_j=\mathbf{0}\) for \(j<s\), since the eigenvalues are distinct and hence \(\lambda_j-\lambda_s\ne0\). The original relation then gives \(\mathbf{v}_s=\mathbf{0}\). Thus a basis chosen in each eigenspace has an independent union.
If \(\mathbf{A}\) is diagonalizable, its eigenvector basis contains exactly \(\dim E_{\lambda_j}\) basis vectors belonging to each \(E_{\lambda_j}\): writing any vector of \(E_{\lambda_j}\) in the eigenvector basis and applying \(\mathbf{A}-\lambda_j\mathbf{I}_n\) forces the coefficients of vectors with other eigenvalues to be zero. In this diagonal representation, the characteristic polynomial has one factor for each basis vector, so the number of entries equal to \(\lambda_j\) is \(\dim E_{\lambda_j}\). Similarity preserves the characteristic polynomial, and for a triangular matrix that number is \(m_j\) by the preceding theorem. Hence \(\dim E_{\lambda_j}=m_j\). Conversely, if all these dimensions equal \(m_j\), choose a basis of each eigenspace. Their union is independent by the argument above, and it contains \(\sum_jm_j=n\) vectors because the characteristic polynomial has degree \(n\). It is therefore a basis of \(\mathbb{R}^n\), so \(\mathbf{A}\) is diagonalizable. \(\square\)
Finding Eigenvectors When the Diagonal Entries Are Distinct
Worked Example: An Upper Triangular Matrix With Three Distinct Eigenvalues
Diagonalize \(\mathbf{A}=\begin{bmatrix}1&1&0\\0&2&1\\0&0&3\end{bmatrix}\). The diagonal entries give the distinct eigenvalues \(1,2,3\), each with algebraic multiplicity \(1\).
For \(\lambda=1\), the system \((\mathbf{A}-\mathbf{I}_3)\mathbf{x}=\mathbf{0}\) gives \(y=0\), \(y+z=0\), and \(2z=0\). Thus \(z=0\), \(y=0\), and \(x\) is free; choose \(\mathbf{v}_1=(1,0,0)^{\top}\). For \(\lambda=2\), the equations are \(-x+y=0\), \(z=0\), and \(z=0\), so choose \(\mathbf{v}_2=(1,1,0)^{\top}\). For \(\lambda=3\), the equations are \(-2x+y=0\) and \(-y+z=0\); taking \(x=1\) gives \(\mathbf{v}_3=(1,2,2)^{\top}\).
Substitution checks the eigenvector equations: \(\mathbf{A}\mathbf{v}_1=(1,0,0)^{\top}=1\mathbf{v}_1\), \(\mathbf{A}\mathbf{v}_2=(2,2,0)^{\top}=2\mathbf{v}_2\), and \(\mathbf{A}\mathbf{v}_3=(3,6,6)^{\top}=3\mathbf{v}_3\). Form \(\mathbf{P}\) from these columns and \(\mathbf{D}\) from their eigenvalues:
The matrix \(\mathbf{P}\) is triangular with determinant \(2\), so its columns form a basis. Multiplying gives \(\mathbf{A}\mathbf{P}=\begin{bmatrix}1&2&3\\0&2&6\\0&0&6\end{bmatrix}\) and \(\mathbf{P}\mathbf{D}=\begin{bmatrix}1&2&3\\0&2&6\\0&0&6\end{bmatrix}\). Thus \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\), which verifies the diagonalization.
With distinct diagonal entries, each eigenspace has at least one nonzero vector, and the eigenspaces contribute independent vectors. There are \(n\) distinct entries for an \(n\times n\) triangular matrix, so one eigenvector from each eigenspace gives a basis. Repeated entries require more care: the eigenspace for a repeated value must provide enough independent vectors to match its algebraic multiplicity.
Repeated Entries: One Success and One Failure
Worked Example: A Repeated Entry With Enough Eigenvectors
Consider \(\mathbf{A}=\begin{bmatrix}2&0&1\\0&2&0\\0&0&3\end{bmatrix}\). Its eigenvalues are \(2\) with algebraic multiplicity \(2\), and \(3\) with algebraic multiplicity \(1\). For \(\lambda=2\), the system \((\mathbf{A}-2\mathbf{I}_3)\mathbf{x}=\mathbf{0}\) reduces to \(z=0\), leaving \(x\) and \(y\) free. Choose \(\mathbf{v}_1=(1,0,0)^{\top}\) and \(\mathbf{v}_2=(0,1,0)^{\top}\). For \(\lambda=3\), the equations are \(-x+z=0\) and \(-y=0\), so choose \(\mathbf{v}_3=(1,0,1)^{\top}\).
Direct checks give \(\mathbf{A}\mathbf{v}_1=(2,0,0)^{\top}=2\mathbf{v}_1\), \(\mathbf{A}\mathbf{v}_2=(0,2,0)^{\top}=2\mathbf{v}_2\), and \(\mathbf{A}\mathbf{v}_3=(3,0,3)^{\top}=3\mathbf{v}_3\). The eigenvector matrix and diagonal matrix are
Here \(\det\mathbf{P}=1\), so the eigenvectors form a basis. Multiplication gives \(\mathbf{A}\mathbf{P}=\begin{bmatrix}2&0&3\\0&2&0\\0&0&3\end{bmatrix} =\mathbf{P}\mathbf{D}\). The repeated eigenvalue has a two-dimensional eigenspace, matching its algebraic multiplicity, so the matrix is diagonalizable.
Worked Example: A Repeated Entry With Too Few Eigenvectors
Now consider \(\mathbf{B}=\begin{bmatrix}2&1&0\\0&2&0\\0&0&3\end{bmatrix}\). Its characteristic polynomial is \((2-\lambda)^2(3-\lambda)\). For \(\lambda=2\), the equations from \((\mathbf{B}-2\mathbf{I}_3)\mathbf{x}=\mathbf{0}\) are \(y=0\) and \(z=0\), with \(x\) free. Therefore \(E_2=\operatorname{span}\{(1,0,0)^{\top}\}\) has dimension \(1\), not \(2\). For \(\lambda=3\), the equations are \(-x+y=0\) and \(-y=0\), so \(x=y=0\) and \(z\) is free. Thus \(E_3=\operatorname{span}\{(0,0,1)^{\top}\}\) also has dimension \(1\).
The vectors \((1,0,0)^{\top}\) and \((0,0,1)^{\top}\) do satisfy the eigenvector equations: their images under \(\mathbf{B}\) are respectively \((2,0,0)^{\top}\) and \((0,0,3)^{\top}\). But together they provide only two independent eigenvectors, not the three required for a basis of \(\mathbb{R}^3\). The eigenspace dimension for \(2\) is smaller than its algebraic multiplicity, so \(\mathbf{B}\) is not diagonalizable.
A Reliable Test and a Common Pitfall
For an upper or lower triangular matrix, begin with the diagonal: list the distinct entries and count how many times each occurs. Then solve \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\) for each distinct entry \(\lambda\), recording a basis for \(E_{\lambda}\). Compare each eigenspace dimension with the number of occurrences of \(\lambda\). The matrix is diagonalizable precisely when every comparison matches. If it is, put the chosen eigenvector bases into the columns of \(\mathbf{P}\), in the same order as their eigenvalues on the diagonal of \(\mathbf{D}\), and check \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\).
Common mistake: treating the number of diagonal entries as the number of independent eigenvectors. For \(\mathbf{B}\) above, a wrong calculation would say: “There are three diagonal entries, so there are three eigenvectors and the matrix is diagonalizable.” The diagonal entries give eigenvalues and algebraic multiplicities, not a count of independent eigenvectors. Solving the equations shows \(\dim E_2=1\) and \(\dim E_3=1\), for only two independent eigenvectors in total. The correction is to find the eigenspaces and compare their dimensions with the repeated-entry counts.
The diagonal entries therefore settle the eigenvalue question immediately, while the eigenspaces settle the diagonalization question. A repeated entry may come with a full eigenspace, as in the successful example, or with too few independent eigenvectors, as in the failure example. The distinction is exactly what the multiplicity test measures.
Check Your Understanding
Use the diagonal entries first, then justify any diagonalization decision by finding the eigenspaces.
- What are the eigenvalues and their algebraic multiplicities for \(\begin{bmatrix}1&2&0\\0&1&3\\0&0&4\end{bmatrix}\)?
- For \(\begin{bmatrix}3&0\\1&3\end{bmatrix}\), solve the eigenvector equation for its repeated eigenvalue and decide whether it is diagonalizable.
- For \(\begin{bmatrix}2&0&0\\0&2&0\\1&0&5\end{bmatrix}\), find a basis for each eigenspace and test diagonalizability.
- Explain why distinct diagonal entries of an \(n\times n\) triangular matrix guarantee diagonalizability over \(\mathbb{R}\).
- State the dimension test for each repeated diagonal value of a triangular matrix to be diagonalizable.
