What a Block Partition Records
A matrix can be divided into smaller rectangular pieces without changing its entries or its size. This is useful when different regions of a matrix have different roles, or when a calculation is easier to organize in pieces. A block is simply a submatrix formed by selecting consecutive rows and consecutive columns. The blocks need not be square, and different blocks need not have the same dimensions.
For example, a \(4\times 3\) matrix may be divided after its second row and after its first column. This produces four blocks: one \(2\times1\), one \(2\times2\), one \(2\times1\), and one \(2\times2\). The horizontal and vertical cuts must span the whole matrix; together they determine the block layout.
In a block display, the block positions form a smaller grid. Within a block row, every block has the same number of rows. Within a block column, every block has the same number of columns. These size requirements make the display a genuine partition of the original matrix rather than an arbitrary arrangement of submatrices.
The partition describes where the pieces are; it does not alter the ordinary definition of matrix addition or scalar multiplication. Addition is defined for matrices of the same size, and scalar multiplication is defined by multiplying every entry by the scalar. The block rules below are consequences of those entrywise definitions. In particular, a block calculation must respect the dimensions of corresponding blocks.
Addition with Matching Partitions
Suppose \(\mathbf{A}\) and \(\mathbf{B}\) are both \(m\times n\) matrices, and their row and column cuts agree. Then \(\mathbf{A}_{ij}\) and \(\mathbf{B}_{ij}\) cover exactly the same row and column positions, so they have the same dimensions and can be added. The sum can be displayed using those same cuts.
Proof. Fix any block position \((i,j)\). For any entry position \((p,q)\) inside that block, the entry of \(\mathbf{A}_{ij}+\mathbf{B}_{ij}\) in its local position is \(a_{pq}+b_{pq}\). By the definition of ordinary matrix addition, this is exactly the entry of \(\mathbf{A}+\mathbf{B}\) in position \((p,q)\). This holds for every position in every block, and the common cuts cover all entries of the matrices. Therefore the blocks of the sum are precisely the sums of the corresponding blocks. \(\square\)
Worked Example: Adding Four Rectangular Blocks
Partition both matrices after row 2 and column 1. Add corresponding blocks and check the resulting full matrix.
The row cuts are \(0,2,4\), and the column cuts are \(0,1,3\). Thus the top-left blocks have size \(2\times1\), the top-right blocks \(2\times2\), the bottom-left blocks \(2\times1\), and the bottom-right blocks \(2\times2\). The block additions are
Putting the four resulting blocks back in their original positions gives
Check: Adding the original matrices entry by entry gives first row \((1+4,\,2+(-2),\,0+1)=(5,0,1)\), second row \((3+(-1),\,1+5,\,4+0)=(2,6,4)\), third row \((2+3,\,-1+2,\,5+(-1))=(5,1,4)\), and fourth row \((0+6,\,3+0,\,2+4)=(6,3,6)\). These are exactly the rows obtained from the block calculation.
Scalar Multiplication Preserves the Cuts
For a real scalar \(c\), the matrix \(c\mathbf{A}\) is obtained by multiplying every entry of \(\mathbf{A}\) by \(c\). If \(\mathbf{A}\) has been partitioned, each of its blocks is multiplied by \(c\), and the row and column cuts stay in the same positions. This also applies when \(c=0\): every block becomes a zero matrix of its original size. A zero block does not remove the cuts in the partition.
Proof. For an entry \(a_{pq}\) in block \(\mathbf{A}_{ij}\), the corresponding entry of \(c\mathbf{A}_{ij}\) is \(ca_{pq}\). By the definition of scalar multiplication of a matrix, the entry in position \((p,q)\) of \(c\mathbf{A}\) is also \(ca_{pq}\). The cuts have not moved, so these entries remain in block position \((i,j)\). The equality holds for every entry of every block, including when \(c=0\), proving the assertion. \(\square\)
Worked Example: Multiplying a Partitioned Matrix by a Scalar
Use row cuts \(0,1,3\) and column cuts \(0,2,4\), and compute \(-2\mathbf{C}\) block by block.
The first block row contains one row, and the second contains two. Each block column contains two columns. The four blocks and their scalar multiples are
Reassembling the blocks in the same positions gives
Check: Multiplying the rows of \(\mathbf{C}\) by \(-2\) gives \((-2,4,-6,0)\), \((2,-8,-4,-10)\), and \((-6,0,4,-2)\), respectively. This agrees with the reassembled blocks.
Addition and scalar multiplication can also be combined without changing the block positions, provided the matrices being added share a partition. The next result records how the familiar scalar and addition rules appear block by block.
Proof. By the scalar multiplication result, \(a\mathbf{A}\) and \(b\mathbf{B}\) retain the partition of their original matrices. Those partitions agree, so the addition result applies. At block position \((i,j)\), the resulting block is \((a\mathbf{A})_{ij}+(b\mathbf{B})_{ij}=a\mathbf{A}_{ij}+b\mathbf{B}_{ij}\), as required. \(\square\)
Worked Example: A Blockwise Linear Combination
For matrices with the same cuts, compute \(2\mathbf{D}-\mathbf{E}\), using row cuts \(0,1,2\) and column cuts \(0,1,3\).
Each matrix has four blocks: two \(1\times1\) blocks in the first block column, and two \(1\times2\) blocks in the second. Compute the corresponding pieces:
Therefore
Check: The first row is \((2\cdot1-2,\,2\cdot2-(-1),\,2\cdot(-1)-5)=(0,5,-7)\). The second is \((2\cdot0-1,\,2\cdot3-0,\,2\cdot4-(-2))=(-1,6,10)\), confirming the result.
When the Partitions Differ
Two matrices of the same size can still be added even if they were originally displayed with different cuts. What fails is only the direct pairing of blocks: a block from one display might cover several blocks in the other. One remedy is to choose common cuts that include every cut from both layouts. This produces a finer partition in which corresponding blocks occupy the same positions and have matching sizes.
Proof. Collect the row cut positions from both partitions, remove duplicates, and arrange them in increasing order, retaining \(0\) and \(m\). Do the same with the column cut positions, retaining \(0\) and \(n\). Every new row interval lies within one row interval of each original partition, since all original row boundaries are included among the new cuts. The same is true of each new column interval. Thus each refined block is contained in one block of each original layout. The matrices have the same overall size, so their entries on every refined block have the same positions and dimensions. The addition theorem therefore applies to the refined blocks, and their union covers the full sum. \(\square\)
Worked Example: Refining Before Adding
Both matrices below are \(2\times3\). The first is cut after column 1; the second is cut after column 2. The row partition is the same, but the column partitions differ.
The first layout has column intervals \(1\mid 2,3\), and the second has intervals \(1,2\mid 3\). Use both column cuts to refine each layout to \(1\mid 2\mid 3\). In this shared layout, the blocks are \(2\times1\) columns, so the additions are
Putting these columns together gives
Check: The first row adds as \((1+(-1),\,2+0,\,3+2)=(0,2,5)\), and the second as \((4+3,\,5+1,\,6+(-2))=(7,6,4)\). The common refinement has made the block addition well organized, while leaving ordinary matrix addition unchanged.
Common Mistake: Pairing Blocks by Position Alone
Matching block labels or block counts does not guarantee that two blocks can be added. Their sizes must agree, which requires the row and column cuts to align. For instance, the first column of \(\mathbf{F}\) in the previous example is a \(2\times1\) block, while the first two columns of \(\mathbf{G}\) form a \(2\times2\) block. A tempting but invalid step is to add these as though they were corresponding blocks: \(\begin{bmatrix}1\\4\end{bmatrix}+\begin{bmatrix}-1&0\\3&1\end{bmatrix}\). The first matrix is \(2\times1\), and the second is \(2\times2\), so this block addition is not defined.
The correction is to refine the column cuts to \(1\mid2\mid3\), as in the example, and then add only blocks of equal size. Alternatively, one can ignore the block display and add entries in the full matrices. The partition is an organizational tool; it never changes which matrices are compatible for ordinary addition.
When using a partition, check three things before calculating: the original matrices have the same overall size, corresponding blocks cover the same row and column ranges, and the result is reassembled in the original order. Scalar multiplication requires no second matrix to match, but it preserves the cuts—including when the scalar is zero. These checks prevent layout choices from being mistaken for changes to the underlying matrix operation.
Check Your Understanding
Use block dimensions and entrywise checks to justify each answer.
- Partition a \(5\times4\) matrix with row cuts \(0,2,5\) and column cuts \(0,1,4\). State the size of each of its four blocks.
- Two \(3\times4\) matrices have row cuts \(0,1,3\) and column cuts \(0,2,4\). Which block of their sum contains the entry in row 3, column 4?
- If a partitioned matrix is multiplied by \(0\), what happens to each block's entries, dimensions, and position?
- Two layouts of a \(4\times5\) matrix use row cuts \(0,2,4\) and \(0,1,4\), respectively. Give the common row cuts for a refinement.
- Explain why the \(2\times1\) and \(2\times2\) blocks in the common-mistake example cannot be added directly, and describe how to fix the layout.
