Reading a Matrix in Blocks
A large matrix can contain a smaller triangular pattern even when it is not itself triangular entry by entry. In that situation, the matrix can be divided into rectangular sections called blocks. If the blocks along the diagonal are square and the blocks on one side of that diagonal are zero, cofactor expansion shows that the determinant is just the product of the determinants of the diagonal blocks. The entries in the remaining off-diagonal block do not change that product.
Work over \(\mathbb{R}\). For a matrix divided into two block rows and two block columns, a block upper-triangular matrix has the form below, where \(\mathbf{A}\) is \(m\times m\), \(\mathbf{D}\) is \(n\times n\), and \(\mathbf{B}\) is \(m\times n\). The zero block has \(n\) rows and \(m\) columns, so the full matrix is square.
A block lower-triangular matrix instead has a zero block above the diagonal blocks:
where \(\mathbf{C}\) is \(n\times m\). In both forms, the diagonal blocks \(\mathbf{A}\) and \(\mathbf{D}\) must be square. The off-diagonal block need not be square, and its entries need not be zero.
The determinant of a square matrix can be expanded along any row or column as a sum of signed entries times the determinants of their corresponding minors. We use that cofactor expansion to prove both cases. The proof does not assume that either diagonal block is invertible, so the rule applies whether or not either block has determinant zero.
Why the Upper-Block Rule Holds
Proof. We prove the upper-triangular formula by induction on \(m+n\), the size of the full matrix. For \(m=1\), the first column of \(\mathbf{M}\) has its only possibly nonzero entry \(a_{11}\) in row 1. Expanding along that column gives \(\det\mathbf{M}=a_{11}\det\mathbf{D}\). Since \(\det\mathbf{A}=a_{11}\), this is the required formula.
Now suppose \(m>1\), and suppose the formula holds for block upper-triangular matrices of smaller total size. The first column of \(\mathbf{M}\) has zero entries in rows \(m+1,\ldots,m+n\). Its cofactor expansion therefore has terms only for \(i=1,\ldots,m\). After deleting row \(i\) and column 1, the remaining matrix is block upper-triangular, with diagonal blocks the \((m-1)\times(m-1)\) minor \(\mathbf{A}_{i1}\) and \(\mathbf{D}\). By the induction hypothesis, that minor of the full matrix has determinant \(\det(\mathbf{A}_{i1})\det\mathbf{D}\). Thus
The final equality is the cofactor expansion of \(\det\mathbf{A}\) down its first column. This proves the upper-block formula for all positive \(m,n\). \(\square\)
Why the Lower-Block Rule Holds
Proof. We again use induction on \(m+n\). When \(m=1\), the first row of \(\mathbf{N}\) has only one possibly nonzero entry, \(a_{11}\), in column 1. Expansion along that row gives \(\det\mathbf{N}=a_{11}\det\mathbf{D}=\det\mathbf{A}\det\mathbf{D}\).
For \(m>1\), assume the formula holds for smaller total sizes. The first row of \(\mathbf{N}\) is zero in columns \(m+1,\ldots,m+n\), so its cofactor expansion uses only columns \(j=1,\ldots,m\). Deleting row 1 and column \(j\) leaves a block lower-triangular matrix whose diagonal blocks are the minor \(\mathbf{A}_{1j}\) and \(\mathbf{D}\). The induction hypothesis says this matrix has determinant \(\det(\mathbf{A}_{1j})\det\mathbf{D}\). Therefore
Here the final sum is the cofactor expansion of \(\det\mathbf{A}\) along its first row. This proves the lower-block formula. \(\square\)
Computing with Block-Triangular Matrices
Worked Example: An Upper-Triangular Matrix with a One-by-One Block
Compute the determinant of the block upper-triangular matrix below. The diagonal blocks have sizes \(2\times2\) and \(1\times1\).
The upper-block rule gives the product of the diagonal-block determinants. First, \(\det\begin{bmatrix}2&1\\3&4\end{bmatrix}=2\cdot4-1\cdot3=5\), and the determinant of the one-by-one block is \(6\). Hence \(\det\mathbf{M}=5\cdot6=30\).
Check: Expanding the full matrix along its last row leaves only the entry 6. Its cofactor sign is \((-1)^{3+3}=1\), and the minor is \(\begin{bmatrix}2&1\\3&4\end{bmatrix}\). Thus the direct expansion gives \(6(2\cdot4-1\cdot3)=6(8-3)=30\), as required.
Worked Example: A Lower-Triangular Matrix with Two-by-Two Blocks
Compute the determinant of this block lower-triangular matrix:
The first diagonal block has determinant \(1\cdot5-2\cdot3=5-6=-1\). The second has determinant \(2\cdot3-(-1)\cdot1=6+1=7\). Therefore \(\det\mathbf{N}=(-1)\cdot7=-7\).
Check: Expand the full matrix along its first row. The cofactor for the entry 1 in column 1 has minor \(\begin{bmatrix}5&0&0\\0&2&-1\\1&1&3\end{bmatrix}\), whose determinant is \(5(2\cdot3-(-1)\cdot1)=35\). The cofactor for the entry 2 in column 2 has minor \(\begin{bmatrix}3&0&0\\4&2&-1\\-2&1&3\end{bmatrix}\), whose determinant is \(3(2\cdot3-(-1)\cdot1)=21\). The column-2 cofactor has a minus sign, so the full determinant is \(1\cdot35-2\cdot21=35-42=-7\). This agrees with the block calculation.
Worked Example: Nonzero Off-Diagonal Block, Two-by-Two Diagonal Blocks
For an upper-block matrix, take
Its diagonal blocks are \(\mathbf{A}=\begin{bmatrix}1&2\\2&5\end{bmatrix}\) and \(\mathbf{D}=\begin{bmatrix}3&1\\4&2\end{bmatrix}\). Their determinants are \(1\cdot5-2\cdot2=1\) and \(3\cdot2-1\cdot4=2\). The block rule gives \(\det\mathbf{M}=1\cdot2=2\), regardless of the entries \(-1,3,2,0\) in the upper-right block.
Check: Expand along the last row. The entry 4 in column 3 has sign \((-1)^{4+3}=-1\); its minor is \(\begin{bmatrix}1&2&3\\2&5&0\\0&0&1\end{bmatrix}\) and has determinant \(1(1\cdot5-2\cdot2)=1\). The entry 2 in column 4 has sign \(+1\); its minor is \(\begin{bmatrix}1&2&-1\\2&5&2\\0&0&3\end{bmatrix}\) and has determinant \(3(1\cdot5-2\cdot2)=3\). The expansion gives \(-4\cdot1+2\cdot3=-4+6=2\), confirming the result.
Common Pitfalls and Useful Extensions
Do not add the diagonal-block determinants. In the first example, adding the diagonal-block determinants would give \(5+6=11\), which is not the determinant. The rule multiplies them: \(5\cdot6=30\).
Check the zero-block position and block sizes. The product rule applies to the stated block upper- or lower-triangular form. It is not enough for a matrix to be divided into four rectangles: the diagonal blocks must be square, and the appropriate off-diagonal block must be zero. An arbitrary block matrix with no such zero pattern is not covered by this rule.
The product formula also handles singular diagonal blocks. For example, if \(\det\mathbf{A}=0\), then the formula gives determinant zero for the full block-triangular matrix. No division by a block determinant appears in the proof, and no block inverse is needed.
A matrix with three or more square diagonal blocks and zeros on one side of the block diagonal can be handled by applying the two-block rule repeatedly. For instance, a block upper-triangular matrix with three diagonal blocks has determinant \(\det\mathbf{A}_1\det\mathbf{A}_2\det\mathbf{A}_3\): first group the last two diagonal blocks into one square block, then apply the rule again to that block. This is useful when a matrix has several natural groups of coordinates, but the same requirements remain: the diagonal blocks are square and the blocks below (or above) the block diagonal are zero.
Check Your Understanding
For each problem, identify the diagonal blocks before applying any determinant rule.
- Compute the determinant of \(\begin{bmatrix}3&-2&4\\0&5&1\\0&0&-2\end{bmatrix}\) by viewing it as a two-block upper-triangular matrix.
- For a block lower-triangular matrix with diagonal blocks \(\begin{bmatrix}2&1\\1&1\end{bmatrix}\) and \(\begin{bmatrix}0&3\\2&4\end{bmatrix}\), find the determinant of the full matrix.
- Does the block upper-triangular rule apply if the lower-left block is nonzero? State which condition is missing.
- Explain why the upper-block proof expands along the first column, while the lower-block proof expands along the first row.
- If one diagonal block of a block-triangular matrix has determinant zero, what does the product rule say about the determinant of the full matrix?
