11.3.1 Example of Matrix-Matrix Multiplication via Linear Combination of Rows

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Modify Rows of a Matrix

Let $A \in R^{4 \times 3}$ . Use 11.3 Matrix-Matrix Multiplication via Linear Combination of Rows to multiply row $i$ by $2^{3 - i}$ .

\begin{proof}[Solution.]
Let's begin by finding our algebraic worker $D$ to multiply each row $i$ by $2^{3 - i}$ . To modify each row, we use a 8.5 Diagonal Matrix. We can get the diagonal matrix we need by multiplying our constraint with each row of the 8.6 Identity Matrix.

\begin{aligned} [D]_{4 \times 4} & = 2^{3 - i} \cdot I (i, :) \\ = {[\begin{array}{c} 2^{3 - 1} & 0 & 0 & 0 \\ 0 & 2^{3 - 2} & 0 & 0 \\ 0 & 0 & 2^{3 - 3} & 0 \\ 0 & 0 & 0 & 2^{3 - 4} \end{array}]}_{4 \times 4} \\ = {[\begin{array}{c} 4 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0.5 \end{array}]}_{4 \times 4} \end{aligned}

We use a $4 \times 4$ matrix because we want to modify 4 rows of $A$ while still maintaining same output dimensions as the input. Recall that the inner dimensions must agree.

B = {[\begin{matrix} 4 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0.5 \end{matrix}]}_{4 \times 4} \cdot {[\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ a_{41} & a_{42} & a_{43} \end{matrix}]}_{4 \times 3}

Now, we can compute the rows of $B$ .

\begin{aligned} B (1, :) & = D (1, :) A \\ = {[\begin{array}{c} 4 & 0 & 0 & 0 \end{array}]}_{1 \times 4} {[\begin{array}{c} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ a_{41} & a_{42} & a_{43} \end{array}]}_{4 \times 3} \\ = 4 [\begin{array}{c} a_{11} & a_{12} & a_{13} \end{array}] + 0 [\begin{array}{c} a_{21} & a_{22} & a_{23} \end{array}] + 0 [\begin{array}{c} a_{31} & a_{32} & a_{33} \end{array}] + 0 [\begin{array}{c} a_{41} & a_{42} & a_{43} \end{array}] \\ = 4 [\begin{array}{c} a_{11} & a_{12} & a_{13} \end{array}] \end{aligned}

We see that the first row of the product is 4 times the first row of matrix $A$ . Let's continue on to row 2 of $B$ .

\begin{aligned} B (2, :) & = D (2, :) A \\ = {[\begin{array}{c} 0 & 2 & 0 & 0 \end{array}]}_{1 \times 4} {[\begin{array}{c} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ a_{41} & a_{42} & a_{43} \end{array}]}_{4 \times 3} \\ = 0 [\begin{array}{c} a_{11} & a_{12} & a_{13} \end{array}] + 2 [\begin{array}{c} a_{21} & a_{22} & a_{23} \end{array}] + 0 [\begin{array}{c} a_{31} & a_{32} & a_{33} \end{array}] + 0 [\begin{array}{c} a_{41} & a_{42} & a_{43} \end{array}] \\ = 2 [\begin{array}{c} a_{21} & a_{22} & a_{23} \end{array}] \end{aligned}

Again, we see that the second row of $B$ is two times the second row of $A$ . Let's continue to row 3.

\begin{aligned} B (3, :) & = D (3, :) A \\ = {[\begin{array}{c} 0 & 0 & 1 & 0 \end{array}]}_{1 \times 4} {[\begin{array}{c} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ a_{41} & a_{42} & a_{43} \end{array}]}_{4 \times 3} \\ = 0 [\begin{array}{c} a_{11} & a_{12} & a_{13} \end{array}] + 0 [\begin{array}{c} a_{21} & a_{22} & a_{23} \end{array}] + 1 [\begin{array}{c} a_{31} & a_{32} & a_{33} \end{array}] + 0 [\begin{array}{c} a_{41} & a_{42} & a_{43} \end{array}] \\ = [\begin{array}{c} a_{31} & a_{32} & a_{33} \end{array}] \end{aligned}

Let's finish with row 4.

\begin{aligned} B (4, :) & = D (4, :) A \\ = {[\begin{array}{c} 0 & 0 & 0 & 0.5 \end{array}]}_{1 \times 4} {[\begin{array}{c} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ a_{41} & a_{42} & a_{43} \end{array}]}_{4 \times 3} \\ = 0 [\begin{array}{c} a_{11} & a_{12} & a_{13} \end{array}] + 0 [\begin{array}{c} a_{21} & a_{22} & a_{23} \end{array}] + 0 [\begin{array}{c} a_{31} & a_{32} & a_{33} \end{array}] + 0.5 [\begin{array}{c} a_{41} & a_{42} & a_{43} \end{array}] \\ = 0.5 [\begin{array}{c} a_{41} & a_{42} & a_{43} \end{array}] \end{aligned}

There seems to be a pattern. For every row of $B$ , we see that it is just multiplied by the corresponding row of $D$ . Let's make a conjecture about this pattern.

Conjecture

Let diagonal matrix $D \in R^{n \times n}$ and $A \in R^{n \times m}$ . When multiplying matrix $A$ on the left by $D$ , we see

\begin{aligned} B (i, :) & = D (i, :) A \\ = D_{1} (c_{1}) D_{2} (c_{2}) \dots D_{i} (c_{i}) \end{aligned}

\end{proof}