Ch02. Matrix Algebra

2.4 Partitioned Matrices

Partitioned Matrices

• A key feature of our work with matrices has been the ability to regard matrix $A$ as a list of column vectors rather than just a rectangular array of numbers.
• This point of view has been so useful that we wish to consider other partitions of $A$, indicated by horizontal and vertical dividing rules, as in Example 1 on the next.

Partition : 분할. ~ to divide matrix into multiple blocks

• to simplify matrix.
• to focus on the impotant part of matrix.

Factorization : 분해.

Example 1:

The matrix

$$A =\left[ \begin{array}{ccc|cc|c} 3 & 0 & -1 & 5 & 9 & -2 \\ -5 & 2 & 4 & 0 &-3 & 1 \\ \hline -8 & -6 & 3 & 1 & 7 &-4 \end{array} \right]$$

can also be written as the $2 \times 3$ partioned (or block) matrix

$$A = \begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \end{bmatrix}$$

Whose entries are the blocks (or submatrices)

$$A_{11}= \begin{bmatrix} 3 & 0 & -1 \\ -5 & 2 & 4 \end{bmatrix}, A_{12}= \begin{bmatrix} 5 & 9 \\ 0 & -3 \end{bmatrix}, A_{13}= \begin{bmatrix} -2 \\ 1 \end{bmatrix} \\ A_{21} = \begin{bmatrix} -8 & -6 & 3 \end{bmatrix}, A_{22} = \begin{bmatrix} 1 & 7 \end{bmatrix}, A_{23} = \begin{bmatrix} -4 \end{bmatrix}$$

Addition and Scalar Multiplication

• If matrices $A$ and $B$ are the same size and are partitioned in exactly the same way, then it is natural to make the same partition of the ordinary matrix sum $A+B$
• In this case, each block of $A+B$ is the (matrix) sum of the corresponding blocks of $A$ and $B$.
• Multiplication of a partitioned matrix by a scalar is also computed block by block.

Multiplication of Partitioned Matrices

• Partitioned matrices can be multiplied by the usual row–column rule as if the block entries were scalars, provided that for a product $AB$, the column partition of $A$ matches the row partition of $B$

Example 3:

Let

$$A =\left[ \begin{array}{ccc|cc} -2 & -3 & 1 & 0 & -4 \\ 1 & 5 & 2 & 3 & -1 \\ \hline 0 & 4 & -2 & 7 & -1 \end{array} \right] = \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix} ,$$ $$B= \left[ \begin{matrix} 6 & 4 \\ -2 & 1 \\ -3 & 7 \\ \hline -1 & 3 \\ 5 & 2 \end {matrix} \right] = \begin{bmatrix} B_1 \\ B_2 \end{bmatrix}$$

• The 5 columns of $A$ are partitioned into a set of 3 columns and then a set of 2 columns. The 5 rows of $B$ are partitioned in the same way –into a set of 3 rows and then a set of 2 rows.

• We say that the partitions of $A$ and $B$ are conformable for block multiplication. It can be shown that the ordinary product $AB$ can be written as $$AB = \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix} \begin{bmatrix} B_1 \\ B_2 \end{bmatrix} = \begin{bmatrix} A_{11}B_{1} + A_{12}B_{2} \\ A_{21}B_{1} + A_{22}B_{2} \end{bmatrix} = \left[ \begin{array} -5 & 4 \\ -6 & 2 \\ \hline 2 & 1 \end{array} \right]$$

• It is important for each smaller product in the expression for $AB$ to be written with the submatrix from $A$ on the left (A로부터 나온 block들도 A처럼 왼쪽에 위치.), since matrix multiplication is not commutative.

• For instance,

$$A_{11}B_1 = \begin{bmatrix} 2 & -3 & 1 \\ 1 & 5 & -2 \end{bmatrix} \begin{bmatrix} 6 & 4 \\ -2 & 1 \\ -3 & 7 \end{bmatrix} = \begin{bmatrix} 15 & 12 \\ 2 & -5 \end{bmatrix} \\ A_{12}B_2 = \begin{bmatrix} 0 & -4 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} -1 & 3 \\ 5 & 2 \end{bmatrix} = \begin{bmatrix} -20 & -8 \\ -8 & 7 \end{bmatrix}$$

• Hence the top block in $AB$ is

$$A_{11}B_1+A_{12}B_1 = \begin{bmatrix} 15 & 12 \\ 2 & -5 \end{bmatrix} + \begin{bmatrix} -20 & -8 \\ -8 & 7 \end{bmatrix} = \begin{bmatrix} -5 & 4 \\ -6 & 2 \end{bmatrix}$$

Theorem : Column—Row Expansion of $AB$

If $A$ is $m \times n$ and $B$ is $n \times p$, then

$$\begin{align} AB &= \begin{bmatrix} \text{col}_{1}(A) & \text{col}_2(A) & \cdots & \text{col}_n(A) \end{bmatrix} \begin{bmatrix} \text{row}_1(B) \\ \text{row}_2(B) \\ \vdots \\ \text{row}_n(B) \end{bmatrix} \\ &= \text{col}_1(A)\text{row}_1(B) + \cdots + \text{col}_n(A) \text{row}_n(B) \tag{1} \end{align}$$

Proof

• For each row index $i$ and column index $j$, the $(i, j)$-entry in $\text{col}_k(A) \text{row}_k{B}$ is the product of $a_{ik}$ from $\text{col}_k(A)$ and $b_{kj}$ from $\text{row}_k(B)$.
• Hence the $(i,j)$-entry in the sum shown in equation (1) is $$\begin{matrix} a_{i1}b_{1j} & + & a_{i2}b_{2j} & + & \cdots & + & a_{in}b_{nj} \\ (k=1) & & (k=2) & & & & (k=n) \end{matrix}$$
• This sum is also the $(i,j)$-entry in $AB$, by the row—column rule.

Inverses of Partioned Matrices

The next example illustrates calculations involving inverses and partitioned matrices.

Example 5:

A matrix of the form $$A=\begin{bmatrix} A_{11} & A_{12} \\ 0 & A_{22} \end{bmatrix}$$ is said to be block upper triangular. Assume that $A_{11}$ is $p \times p$, $A_{22}$ is $q \times q$, and $A$ is invertible. Find a formula for $A^{-1}$.

Solution

Denote $A^{-1}$ by $B$ and partition $B$ so that,

$$\begin{bmatrix} A_{11} & A_{12} \\ 0 & A_{22} \end{bmatrix} \begin{bmatrix} B_{11} & B_{12} \\ B_{21} & B_{22} \end {bmatrix} = \begin{bmatrix} I_p & 0 \\ 0 & I_q \end{bmatrix} \tag{2}$$

• This matrix equation provides four equations that will lead to the unknown blocks, $B_{11} \cdots, B_{22}$. Compute the product on the left side of equation (2), and equate each entry with the corresponding block in the identity matrix on the right.
• That is, set

$${A_{11}B_{11} + A_{12}B_{21} = I_p}\quad \tag{3}$$ $${A_{11}B_{12} + A_{12}B_{22} = 0}\quad \tag{4}$$ $${A_{22}B_{21} = 0}\quad \tag{5}$$ $${A_{22}B_{22} = I_q}\quad \tag{6}$$

• By itself, equation (6) does not show that $A_{22}$ is invertible. However, since $A_{22}$ is square, the Invertible Matrix Theorem and (6) together show that $A_{22}$ is invertible and $B_{22}=A_{22}^{-1}$.
• Next, left-multiply both sides of (5) by $A_{22}^{-1}$ and obtain $$B_{21}= A_{22}^{-1}0=0$$ .
• So that (3) simplifies to $$A_{11}B_{11}+0 = I_p$$ .
• Since $A_{11}$ is square, this shows that $A_{11}$ is invertible and $B_{22}=A_{22}^{-1}$. Finally, use these results with (4) to find that

$$A_{11}B_{12} = -A_{12}B_{22} = -A_{12}A_{22}^{-1} \text{ and } B_{12}=-A_{11}^{-1}A_{12}A_{22}^{-1}$$

• Thus

$$A^{-1}=\begin{bmatrix} A_{11} & A_{12} \\ 0 & A_{22} \end{bmatrix} = \begin{bmatrix} A_{11}^{-1} & -A_{11}^{-1}A_{12}A_{22}^{-1} \\ 0 & A_{22}^{-1} \end {bmatrix} \\ \blacksquare$$

• A block diagonal matrix is a partitioned matrix with zero blocks off the main diagonal (of blocks). Such a matrix is invertible if and only if each block on the diagonal is invertible.