Diagonal Entries and Characteristic Polynomials
The characteristic polynomial packages the eigenvalues of a matrix into one polynomial. For a triangular matrix, its computation becomes especially direct: the diagonal entries determine the polynomial, while entries above the diagonal do not change it. This fact is useful beyond matrices that are already triangular. Over \(\mathbb{C}\), every square matrix is similar to an upper-triangular matrix, so the diagonal of a triangular form accounts for all its eigenvalues, including repeated roots.
Throughout, matrices have size \(n\times n\), and the characteristic polynomial uses the convention \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_n)\). The scalar \(\lambda\) may be real or complex, depending on the matrix. An eigenvalue is a scalar for which \(\mathbf{A}-\lambda\mathbf{I}_n\) is singular; equivalently, there is a nonzero vector \(\mathbf{v}\) satisfying \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). The algebraic multiplicity of an eigenvalue is its multiplicity as a root of \(p_{\mathbf{A}}\).
Proof. Subtracting \(\lambda\mathbf{I}_n\) changes only the diagonal, so \(\mathbf{T}-\lambda\mathbf{I}_n\) is upper triangular with diagonal entries \(d_j-\lambda\). By the permutation formula for a determinant, a term indexed by a permutation \(\sigma\) is a signed product of entries from positions \((i,\sigma(i))\). Such a product is zero unless every selected position is on or above the diagonal, which requires \(\sigma(i)\ge i\) for every \(i\). Since \(\sum_i\sigma(i)=\sum_i i\), these inequalities can all hold only with \(\sigma(i)=i\) for every \(i\). Thus the identity permutation supplies the only possibly nonzero term, and that term is \(\prod_{j=1}^{n}(d_j-\lambda)\). For a lower-triangular matrix, a nonzero term requires \(\sigma(i)\le i\) for every \(i\); the same sum argument again leaves only the identity permutation. \(\square\)
The factorization gives the roots directly: the eigenvalues of a triangular matrix are its diagonal entries. If an entry occurs more than once, it contributes a repeated factor, so its number of occurrences is its algebraic multiplicity. The order of the diagonal entries does not matter to the product, but retaining their repetitions does.
Worked Example: Reading Roots and Multiplicities from the Diagonal
Find the characteristic polynomial, eigenvalues, and algebraic multiplicities of
The matrix is upper triangular, with diagonal entries \(2,2,-1\). Therefore
The roots are \(2\) and \(-1\). The root \(2\) occurs twice, so \(\operatorname{am}(2)=2\); the root \(-1\) occurs once, so \(\operatorname{am}(-1)=1\).
Check: The diagonal entries of \(\mathbf{T}-\lambda\mathbf{I}_3\) are \(2-\lambda,2-\lambda,-1-\lambda\), and its determinant is their product by the triangular determinant formula. The polynomial has degree \(3\), as required for a \(3\times3\) matrix.
Why Triangular Forms Describe Other Matrices
The earlier material on complex triangularization establishes that every square matrix over \(\mathbb{C}\) is similar over \(\mathbb{C}\) to an upper-triangular matrix. Similarity means that \(\mathbf{B}=\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\) for an invertible matrix \(\mathbf{P}\). To transfer a triangular matrix’s characteristic polynomial back to the original matrix, we need the fact that similarity preserves the characteristic polynomial.
Proof. Suppose \(\mathbf{B}=\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\) with \(\mathbf{P}\) invertible. Then \(\mathbf{I}_n=\mathbf{P}^{-1}\mathbf{I}_n\mathbf{P}\), so \(\mathbf{B}-\lambda\mathbf{I}_n=\mathbf{P}^{-1}(\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{P}\). Using the determinant product rule and \(\det(\mathbf{P}^{-1})\det(\mathbf{P})=1\), we obtain \[ \det(\mathbf{B}-\lambda\mathbf{I}_n) =\det(\mathbf{P}^{-1})\det(\mathbf{A}-\lambda\mathbf{I}_n)\det(\mathbf{P}) =\det(\mathbf{A}-\lambda\mathbf{I}_n). \] This is the claimed equality of characteristic polynomials. \(\square\)
Proof. By similarity invariance, \(p_{\mathbf{A}}=p_{\mathbf{T}}\). By the triangular determinant formula, \(p_{\mathbf{T}}(\lambda)=\prod_{j=1}^{n}(d_j-\lambda)\). A scalar \(\lambda\) is an eigenvalue exactly when \(\det(\mathbf{A}-\lambda\mathbf{I}_n)=0\), which is exactly when this product is zero. A product of these linear factors vanishes precisely at the diagonal values. For any fixed value \(d\), each occurrence of \(d\) supplies one factor \(d-\lambda\), so the root \(d\) has multiplicity equal to the number of those occurrences. \(\square\)
This conclusion accounts for every eigenvalue over \(\mathbb{C}\), even when the original matrix has no real eigenvalues. It also explains why the diagonal of a triangular form is more than a convenient way to compute a polynomial: it gives the complete eigenvalue list with repetitions. Different triangular forms of the same matrix may arrange the diagonal entries differently, but similarity invariance forces the same characteristic polynomial, so the eigenvalues and their algebraic multiplicities agree.
Worked Applications
Worked Example: A Real Matrix with Complex Eigenvalues
Consider the real matrix \(\mathbf{A}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}\). Find its characteristic polynomial and use a complex triangular form to identify its eigenvalues.
First, compute the polynomial directly:
Over \(\mathbb{C}\), choose \(\mathbf{v}_1=\begin{bmatrix}1\\-i\end{bmatrix}\) and \(\mathbf{v}_2=\begin{bmatrix}1\\i\end{bmatrix}\), and form \(\mathbf{P}=\begin{bmatrix}1&1\\-i&i\end{bmatrix}\). Its determinant is \(2i\ne0\), so it is invertible. Direct multiplication gives \(\mathbf{A}\mathbf{v}_1=\begin{bmatrix}i\\1\end{bmatrix}=i\mathbf{v}_1\) and \(\mathbf{A}\mathbf{v}_2=\begin{bmatrix}-i\\1\end{bmatrix}=-i\mathbf{v}_2\). Thus \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\), where \(\mathbf{D}=\begin{bmatrix}i&0\\0&-i\end{bmatrix}\), and hence \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}=\mathbf{D}\).
The diagonal factors give \((i-\lambda)(-i-\lambda)=\lambda^2+1\), since \(i(-i)=1\) and the two terms linear in \(\lambda\) cancel. The eigenvalues are \(i\) and \(-i\), each with algebraic multiplicity \(1\).
Check: Substitution gives \(\mathbf{A}\mathbf{v}_1=i\mathbf{v}_1\) and \(\mathbf{A}\mathbf{v}_2=-i\mathbf{v}_2\), as computed above. The factors from the triangular form reproduce the directly computed characteristic polynomial.
For a real matrix, treating it as a complex matrix does not alter its characteristic polynomial; it only allows nonreal roots to be included. Such roots occur in conjugate pairs with equal algebraic multiplicity. Indeed, a real-coefficient polynomial satisfies \(p(\overline{\lambda})=\overline{p(\lambda)}\). Conjugating a factorization over \(\mathbb{C}\) therefore pairs each factor for a nonreal root with the same number of factors for its conjugate.
Worked Example: Repeated Roots Need Not Give Many Eigenvectors
Let \(\mathbf{T}=\begin{bmatrix}3&1&0\\0&3&0\\0&0&-2\end{bmatrix}\). Find the algebraic multiplicities and compare the eigenspace for \(3\) with its algebraic multiplicity.
The diagonal formula gives \[ p_{\mathbf{T}}(\lambda)=(3-\lambda)^2(-2-\lambda). \] Thus \(3\) has algebraic multiplicity \(2\), and \(-2\) has algebraic multiplicity \(1\). To find the eigenvectors for \(3\), solve \((\mathbf{T}-3\mathbf{I}_3)\mathbf{x}=\mathbf{0}\): \[ \begin{bmatrix}0&1&0\\0&0&0\\0&0&-5\end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} =\begin{bmatrix}0\\0\\0\end{bmatrix}. \] The first row gives \(x_2=0\), the third gives \(x_3=0\), and \(x_1\) is free. Therefore the eigenvectors together with \(\mathbf{0}\) form the one-dimensional space \(\{(x_1,0,0)^{\top}:x_1\in\mathbb{R}\}\).
Check: For every \(x_1\), \(\mathbf{T}\begin{bmatrix}x_1\\0\\0\end{bmatrix}=\begin{bmatrix}3x_1\\0\\0\end{bmatrix}=3\begin{bmatrix}x_1\\0\\0\end{bmatrix}\). In this example, the algebraic multiplicity of \(3\) is \(2\), while its eigenspace has dimension \(1\). The repeated factor records root multiplicity, not the number of independent eigenvectors.
Common Mistakes and a Reliable Procedure
Mistake: treating an off-diagonal entry as part of the triangular determinant product. For example, an incorrect calculation might give \(\det\begin{bmatrix}2-\lambda&5\\0&3-\lambda\end{bmatrix}=(2-\lambda)(3-\lambda)-5\). The \(5\) is not multiplied by a nonzero entry in its permutation term: the other selected entry would be the zero below the diagonal. The correct calculation is \[ \det\begin{bmatrix}2-\lambda&5\\0&3-\lambda\end{bmatrix} =(2-\lambda)(3-\lambda)-5\cdot0 =(2-\lambda)(3-\lambda). \] The diagonal-factor rule applies because the matrix is triangular; it is not a rule for arbitrary matrices.
Mistake: listing only distinct eigenvalues when asked for algebraic multiplicity. In the first worked example, the eigenvalue list without repetition is \(\{2,-1\}\), but the characteristic polynomial is \((2-\lambda)^2(-1-\lambda)\). The repeated factor is essential: it records that the algebraic multiplicity of \(2\) is \(2\), not \(1\).
For a complex matrix that is not already triangular, the reasoning can be organized in three steps:
Use complex triangularization to write \(\mathbf{A}=\mathbf{P}\mathbf{T}\mathbf{P}^{-1}\), with \(\mathbf{T}\) upper triangular.
If the diagonal of \(\mathbf{T}\) is \(d_1,\dots,d_n\), then \(p_{\mathbf{T}}(\lambda)=\prod_{j=1}^{n}(d_j-\lambda)\).
Similarity preserves the characteristic polynomial, so the same product is \(p_{\mathbf{A}}(\lambda)\). Count repeated diagonal values to obtain algebraic multiplicities.
The method gives a characteristic polynomial and the complete eigenvalue list without requiring the original matrix to be triangular. When the matrix is real but its triangular form is complex, the factors may be complex; the resulting characteristic polynomial still has real coefficients, and nonreal roots occur with their conjugates.
Check Your Understanding
Use the diagonal-factor formula and similarity invariance where appropriate.
- Find the characteristic polynomial and algebraic multiplicities for an upper-triangular matrix with diagonal entries \(1,-2,1,4\).
- Why does the permutation formula leave only the identity-permutation term for the determinant of an upper-triangular matrix?
- If \(\mathbf{A}\) is similar over \(\mathbb{C}\) to a triangular matrix with diagonal \(i,i,-i,2\), list its eigenvalues and their algebraic multiplicities.
- For a real matrix, explain why a nonreal eigenvalue must be accompanied by its complex conjugate as a root of the characteristic polynomial.
- Give the characteristic polynomial of \(\begin{bmatrix}5&7\\0&5\end{bmatrix}\). Does the repeated root alone determine the dimension of its eigenspace?
