Why the Diagonal Blocks Determine the Polynomial
A block triangular matrix may contain nonzero entries away from its diagonal blocks. Nevertheless, those entries do not change its characteristic polynomial: the polynomial is the product of the characteristic polynomials of the diagonal blocks. The key is the zero block. In the determinant expansion, it rules out every term that would mix the two diagonal blocks.
Throughout this tutorial, matrices have real entries. For a square matrix \(\mathbf{A}\), its characteristic polynomial is \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I})\). The symbol \(\lambda\) is an indeterminate, so the entries of \(\mathbf{A}-\lambda\mathbf{I}\) are polynomials. We will use the determinant expansion over those polynomials.
The diagonal blocks need not be the same size, and the off-diagonal block need not be zero. A matrix with several diagonal blocks is block upper triangular if every block strictly below the diagonal blocks is zero; it is block lower triangular if every block strictly above them is zero. In each case, all diagonal blocks are square.
The Determinant Product Rule
We first establish the determinant fact that makes the characteristic-polynomial formula work. For an \(N\times N\) matrix \(\mathbf{M}\), the determinant expansion is \(\det\mathbf{M}=\sum_{\pi\in S_N}\operatorname{sgn}(\pi)\prod_{i=1}^{N}m_{i,\pi(i)}\), where \(S_N\) is the set of permutations of \(1,\dots,N\), and \(\operatorname{sgn}(\pi)\) is the sign of the permutation.
Proof. First consider the upper block matrix, where \(\mathbf{B}\) is \(m\times m\) and \(\mathbf{D}\) is \(n\times n\). In any nonzero term of the determinant expansion, each of the last \(n\) rows must be paired by the permutation with a column from the last \(n\) columns: entries in those rows and the first \(m\) columns are all zero. Because the permutation uses each column exactly once, the last \(n\) rows use all the last \(n\) columns, and the first \(m\) rows use all the first \(m\) columns. Thus every nonzero term comes from a permutation of the first \(m\) indices and a permutation of the last \(n\) indices. The combined permutation has sign equal to the product of their signs: it has no inversions between the two groups, since every index in the first group precedes every index in the second. Therefore the determinant expansion separates into the product of the expansion for \(\det\mathbf{B}\) and the expansion for \(\det\mathbf{D}\). For the lower block matrix, the first \(m\) rows must use the first \(m\) columns, because their entries in the last \(n\) columns are zero. The same permutation and sign argument separates its expansion into \(\det\mathbf{B}\det\mathbf{D}\). \(\square\)
The proof uses only the zero block and the determinant expansion. It does not require \(\mathbf{C}\) to be zero, and it does not require either diagonal block to be invertible.
Proof. For two diagonal blocks, subtracting \(\lambda\) from the diagonal entries preserves the block triangular form, and gives diagonal blocks \(\mathbf{B}-\lambda\mathbf{I}\) and \(\mathbf{D}-\lambda\mathbf{I}\). The determinant product rule therefore gives \(\det(\mathbf{A}-\lambda\mathbf{I})=\det(\mathbf{B}-\lambda\mathbf{I})\det(\mathbf{D}-\lambda\mathbf{I})\), which is \(p_{\mathbf{A}}(\lambda)=p_{\mathbf{B}}(\lambda)p_{\mathbf{D}}(\lambda)\). For more than two diagonal blocks, group the first block against all the remaining blocks. The grouped matrix is still two-block triangular, and the second diagonal block is itself block triangular. Applying the two-block result repeatedly gives the product of the characteristic polynomials of all diagonal blocks. The argument works for both upper and lower block triangular forms. \(\square\)
Because the eigenvalues of a square matrix are the roots of its characteristic polynomial, this formula also shows that an eigenvalue of a block triangular matrix must be a root of the characteristic polynomial of at least one diagonal block. The off-diagonal blocks can be nonzero without changing that polynomial.
Applying the Formula
Worked Example: A Two-by-Two Block and a Scalar Block
Find the characteristic polynomial of \(\mathbf{A}=\begin{bmatrix}1&2&5\\0&3&-1\\0&0&4\end{bmatrix}\). Use the first two rows and columns as one diagonal block, so the upper-right block is nonzero.
The diagonal blocks are \(\mathbf{B}=\begin{bmatrix}1&2\\0&3\end{bmatrix}\) and \(\mathbf{D}=\begin{bmatrix}4\end{bmatrix}\). Subtracting \(\lambda\mathbf{I}\) from each gives
The first matrix is upper triangular, so its determinant is the product of its diagonal entries. The block product rule then gives
Check: Direct expansion of \(\mathbf{A}-\lambda\mathbf{I}_3\) along its last row gives \((4-\lambda)\det\begin{bmatrix}1-\lambda&2\\0&3-\lambda\end{bmatrix}=(4-\lambda)(1-\lambda)(3-\lambda)\), the same polynomial. At \(\lambda=0\), it gives \(\det\mathbf{A}=1\cdot3\cdot4=12\), agreeing with the constant term.
Worked Example: A Lower Block Triangular Matrix
Find the characteristic polynomial of \(\mathbf{A}=\begin{bmatrix}2&0&0\\3&1&2\\5&0&4\end{bmatrix}\). Partition it into a \(1\times1\) block and a \(2\times2\) block.
This is block lower triangular, with diagonal blocks \(\begin{bmatrix}2\end{bmatrix}\) and \(\begin{bmatrix}1&2\\0&4\end{bmatrix}\). Their shifted matrices are
The determinant product rule, followed by the \(2\times2\) triangular determinant calculation, gives
Check: Expanding \(\det(\mathbf{A}-\lambda\mathbf{I}_3)\) along the first row leaves \((2-\lambda)\det\begin{bmatrix}1-\lambda&2\\0&4-\lambda\end{bmatrix}\), which is precisely the factored expression. Also, at \(\lambda=0\), \(\det\mathbf{A}=2(1\cdot4-2\cdot0)=8\), matching the constant term.
Worked Example: Three Scalar Diagonal Blocks
Find the characteristic polynomial of \(\mathbf{T}=\begin{bmatrix}-1&2&7\\0&2&-3\\0&0&5\end{bmatrix}\). Here each diagonal block is a scalar, and there are nonzero entries in both off-diagonal positions above the diagonal.
The three diagonal blocks are \([-1]\), \([2]\), and \([5]\). The theorem gives
Check: Direct expansion along the last row of \(\mathbf{T}-\lambda\mathbf{I}_3\) gives \((5-\lambda)\det\begin{bmatrix}-1-\lambda&2\\0&2-\lambda\end{bmatrix}=(5-\lambda)(-1-\lambda)(2-\lambda)\). At \(\lambda=0\), the determinant is \((-1)\cdot2\cdot5=-10\), matching the constant term.
What the Block Structure Does—and Does Not—Guarantee
The determinant product rule applies to the whole displayed partition, not merely to any matrix that has been divided into rectangular pieces. One of the off-diagonal blocks must be zero. This condition is what forces the determinant terms to separate. Nonzero entries in the other off-diagonal block are allowed; nonzero entries in both are not covered by the theorem.
Common mistake: applying the rule to a general block matrix. For example, consider \(\mathbf{M}=\begin{bmatrix}1&1\\1&2\end{bmatrix}\), partitioned into four \(1\times1\) blocks. Since neither off-diagonal block is zero, \(\mathbf{M}\) is not block triangular. Incorrectly multiplying the characteristic polynomials of its diagonal blocks would give \((1-\lambda)(2-\lambda)=\lambda^2-3\lambda+2\). But direct calculation gives
The missing \(-1\) comes from the nonzero off-diagonal entries. The block product formula cannot be used here because neither required zero block is present.
A second useful check is the sign of the leading term. For an \(N\times N\) matrix, \(\det(\mathbf{A}-\lambda\mathbf{I}_N)\) has leading term \((-\lambda)^N\). When multiplying the characteristic polynomials of the diagonal blocks, their degrees add to \(N\), and their leading terms multiply to \((-\lambda)^N\). If a proposed product has the wrong degree or leading sign, an arithmetic or convention error has occurred.
| Matrix form | Product rule applies? | Reason |
|---|---|---|
| \(\begin{bmatrix}\mathbf{B}&\mathbf{C}\\\mathbf{O}&\mathbf{D}\end{bmatrix}\) | Yes | Block upper triangular |
| \(\begin{bmatrix}\mathbf{B}&\mathbf{O}\\\mathbf{C}&\mathbf{D}\end{bmatrix}\) | Yes | Block lower triangular |
| \(\begin{bmatrix}\mathbf{B}&\mathbf{C}\\\mathbf{E}&\mathbf{D}\end{bmatrix}\), with both \(\mathbf{C}\) and \(\mathbf{E}\) nonzero | Not from this theorem | Neither off-diagonal block is zero |
When a matrix is block triangular, identify the diagonal blocks first, shift each by \(\lambda\mathbf{I}\) using the convention \(\mathbf{A}-\lambda\mathbf{I}\), and find each block’s determinant. Multiply those polynomials and check the degree and leading term. This keeps the calculation focused on the entries that determine the characteristic polynomial, while preserving the important distinction between a block triangular matrix and an arbitrary block matrix.
Check Your Understanding
For each question, state the block structure you are using and give enough calculation to justify your answer.
- Find the characteristic polynomial of \(\begin{bmatrix}3&4&1\\0&-2&5\\0&0&1\end{bmatrix}\) by identifying its diagonal blocks.
- For \(\begin{bmatrix}1&0&0\\2&2&3\\4&0&-1\end{bmatrix}\), identify a two-block partition and write the characteristic polynomial as a product before expanding.
- State why the determinant product rule applies to \(\begin{bmatrix}\mathbf{B}&\mathbf{C}\\\mathbf{O}&\mathbf{D}\end{bmatrix}\) even when \(\mathbf{C}\ne\mathbf{O}\).
- Compute the characteristic polynomial of \(\begin{bmatrix}1&2\\3&4\end{bmatrix}\) directly, and explain why the diagonal-block product rule does not apply to its \(1\times1\) partition.
- If a block upper triangular matrix has diagonal blocks of sizes \(2\times2\) and \(3\times3\), what is the degree and leading term of the product characteristic polynomial?
