Triangular Matrices And Their Eigenvalues

Why the Diagonal Determines the Eigenvalues

For a general matrix, finding eigenvalues can require computing a characteristic polynomial and then finding its roots. A triangular matrix has a useful simplification: its characteristic polynomial factors directly into terms involving its diagonal entries. This means its eigenvalues can be read off without expanding a determinant or solving a new polynomial equation.

Throughout, let \(\mathbb{F}\) be either \(\mathbb{R}\) or \(\mathbb{C}\), and let \(\mathbf{A}\in\mathbb{F}^{n\times n}\), where \(n\ge 1\). A matrix is upper triangular if its entries below the diagonal are zero, and lower triangular if its entries above the diagonal are zero. A matrix that is both is diagonal. Write its diagonal entries as \(a_{11},a_{22},\dots,a_{nn}\).

The sign convention matters when writing the polynomial. Since \(p_{\mathbf{A}}(\lambda)\) is defined using \(\mathbf{A}-\lambda\mathbf{I}_n\), its factors for a triangular matrix are \(a_{ii}-\lambda\). Some books instead use \(\det(\lambda\mathbf{I}_n-\mathbf{A})\), which has factors \(\lambda-a_{ii}\). These polynomials differ by the constant factor \((-1)^n\), so they have the same roots with the same multiplicities.

The Determinant of a Triangular Matrix

We first justify the determinant calculation that makes the eigenvalue formula work. For an \(n\times n\) matrix \(\mathbf{B}=(b_{ij})\), the determinant can be written as a sum over permutations \(\sigma\) of \(\{1,\dots,n\}\): each term is a signed product \(b_{1,\sigma(1)}b_{2,\sigma(2)}\cdots b_{n,\sigma(n)}\). A term is zero if any entry in its product is zero.

Proof. First suppose that \(\mathbf{B}\) is upper triangular, so \(b_{ij}=0\) whenever \(i>j\). For a term indexed by a permutation \(\sigma\) to be nonzero, every selected entry \(b_{i,\sigma(i)}\) must be nonzero. Upper triangularity therefore requires \(\sigma(i)\ge i\) for every \(i\). But \(\sum_{i=1}^{n}\sigma(i)=\sum_{i=1}^{n}i\), since \(\sigma\) is a permutation. The nonnegative integers \(\sigma(i)-i\) have sum zero, so each one is zero. Thus only the identity permutation can contribute a nonzero term, and that term is \(b_{11}b_{22}\cdots b_{nn}\). For a lower triangular matrix, a nonzero term requires \(\sigma(i)\le i\) for every \(i\). The nonnegative integers \(i-\sigma(i)\) also have sum zero, so again \(\sigma(i)=i\) for each \(i\). The determinant is therefore the diagonal product in either case. \(\square\)

Subtracting \(\lambda\) from each diagonal entry preserves triangularity: \(\mathbf{A}-\lambda\mathbf{I}_n\) has the same zero entries above or below the diagonal as \(\mathbf{A}\). Applying the determinant theorem gives the characteristic-polynomial formula.

Proof. The diagonal entries of \(\mathbf{A}-\lambda\mathbf{I}_n\) are \(a_{ii}-\lambda\), and that matrix is triangular. The determinant theorem gives \(\det(\mathbf{A}-\lambda\mathbf{I}_n)=\prod_{i=1}^{n}(a_{ii}-\lambda)\). For the alternate convention, \(\lambda\mathbf{I}_n-\mathbf{A}=-(\mathbf{A}-\lambda\mathbf{I}_n)\); its diagonal entries are \(\lambda-a_{ii}\), so the same determinant theorem gives the second formula directly. \(\square\)

Reading Off Eigenvalues and Their Multiplicities

The earlier material on the eigenvalue equation and singularity establishes that \(\lambda\) is an eigenvalue of \(\mathbf{A}\) exactly when \(\det(\mathbf{A}-\lambda\mathbf{I}_n)=0\). For a triangular matrix, the factorization now makes this test immediate: the product is zero exactly when at least one factor \(a_{ii}-\lambda\) is zero.

Proof. By the characteristic-polynomial formula, \(p_{\mathbf{A}}(\lambda)=\prod_{i=1}^{n}(a_{ii}-\lambda)\). This product is zero if and only if at least one factor is zero, which happens if and only if \(\lambda=a_{ii}\) for some \(i\). By the eigenvalue-and-singularity criterion, these and only these values are eigenvalues. If a particular value \(\mu\) occurs \(r\) times among \(a_{11},\dots,a_{nn}\), then exactly \(r\) factors in the product equal \(\mu-\lambda\). The other factors do not vanish at \(\lambda=\mu\), so the root has multiplicity \(r\). \(\square\)

This conclusion concerns roots and their multiplicities in the characteristic polynomial. It does not say how many linearly independent eigenvectors there are for a repeated eigenvalue; the diagonal entries alone do not settle that question.

Common Pitfall: Losing the Sign

With the course convention \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_n)\), writing every factor as \(\lambda-a_{ii}\) can give the wrong polynomial, even though it gives the right roots. For example, take the upper triangular matrix \(\mathbf{D}=\begin{bmatrix}1&2&0\\0&2&3\\0&0&4\end{bmatrix}\). A mistaken calculation would write \(p_{\mathbf{D}}(\lambda)=(\lambda-1)(\lambda-2)(\lambda-4)\). But the defined matrix \(\mathbf{D}-\lambda\mathbf{I}_3\) has diagonal entries \(1-\lambda,2-\lambda,4-\lambda\), so the correct calculation is

The incorrect expression has the opposite sign because there are three factors. Both expressions vanish at \(1,2,4\), so the error does not change the eigenvalues; it does change the characteristic polynomial. A reliable check is to apply the definition first: write \(\mathbf{A}-\lambda\mathbf{I}_n\), read its diagonal entries, and only then multiply the factors.

Finally, the diagonal-entry rule requires a triangular matrix. For a matrix that is not triangular, its diagonal entries alone do not in general determine its eigenvalues. The rule works here because triangularity makes the determinant of \(\mathbf{A}-\lambda\mathbf{I}_n\) a product of its diagonal entries, not because every matrix has that determinant formula.