The basic definition of matrix multiplication A B tells us...

The basic definition of matrix multiplication $A B$ tells us to multiply rows of $A$ by columns of $B$. Remarkably, if you suitably interpret the operation, you can also compute $A B$ by multiplying columns of $A$ by rows of $B$ ! Suppose $A$ is an $m \times n$ matrix with columns $\mathbf{c}_1, \ldots, \mathbf{c}_n$. Suppose $B$ is an $n \times p$ matrix with rows $\mathbf{r}_1, \ldots, \mathbf{r}_n$. Then we claim that $$ A B=\mathbf{c}_1 \mathbf{r}_1+\mathbf{c}_2 \mathbf{r}_2+\cdots+\mathbf{c}_n \mathbf{r}_n \text {, } $$ where each summand is a matrix of size $m \times p$. (a) Verify that the particular case $$ \left(\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right)\left(\begin{array}{rr} 0 & -1 \\ 2 & 3 \end{array}\right)=\left(\begin{array}{l} 1 \\ 3 \end{array}\right)\left(\begin{array}{ll} 0 & -1 \end{array}\right)+\left(\begin{array}{l} 2 \\ 4 \end{array}\right)\left(\begin{array}{ll} 2 & 3 \end{array}\right)=\left(\begin{array}{rr} 0 & -1 \\ 0 & -3 \end{array}\right)+\left(\begin{array}{rr} 4 & 6 \\ 8 & 12 \end{array}\right)=\left(\begin{array}{ll} 4 & 5 \\ 8 & 9 \end{array}\right) $$ agrees with the usual method for computing the matrix product. (b) Use this method to compute the matrix products $(i)\left(\begin{array}{rr}-2 & 1 \\ 3 & 2\end{array}\right)\left(\begin{array}{rr}1 & -2 \\ 1 & 0\end{array}\right)$, (ii) $$ \left(\begin{array}{rrr} 1 & -2 & 0 \\ -3 & -1 & 2 \end{array}\right)\left(\begin{array}{rr} 2 & 5 \\ -3 & 0 \\ 1 & -1 \end{array}\right) $$ (iii) $\left(\begin{array}{rrr}3 & -1 & 1 \\ -1 & 2 & 1 \\ 1 & 1 & -5\end{array}\right)\left(\begin{array}{rrr}2 & 3 & 0 \\ 3 & -1 & 4 \\ 0 & 4 & 1\end{array}\right)$, and verify that you get the same answer as that obtained by the traditional method. (c) Explain why (1.13) is a special case of (1.14). (d) Prove that (1.14) gives the correct formula for the matrix product.

The basic definition of matrix multiplication $A B$ tells us to multiply rows of $A$ by columns of $B$. Remarkably, if you suitably interpret the operation, you can also compute $A B$ by multiplying columns of $A$ by rows of $B$ ! Suppose $A$ is an $m \times n$ matrix with columns $\mathbf{c}_1, \ldots, \mathbf{c}_n$. Suppose $B$ is an $n \times p$ matrix with rows $\mathbf{r}_1, \ldots, \mathbf{r}_n$. Then we claim that
$$
A B=\mathbf{c}_1 \mathbf{r}_1+\mathbf{c}_2 \mathbf{r}_2+\cdots+\mathbf{c}_n \mathbf{r}_n \text {, }
$$
where each summand is a matrix of size $m \times p$. (a) Verify that the particular case
$$
\left(\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right)\left(\begin{array}{rr}
0 & -1 \\
2 & 3
\end{array}\right)=\left(\begin{array}{l}
1 \\
3
\end{array}\right)\left(\begin{array}{ll}
0 & -1
\end{array}\right)+\left(\begin{array}{l}
2 \\
4
\end{array}\right)\left(\begin{array}{ll}
2 & 3
\end{array}\right)=\left(\begin{array}{rr}
0 & -1 \\
0 & -3
\end{array}\right)+\left(\begin{array}{rr}
4 & 6 \\
8 & 12
\end{array}\right)=\left(\begin{array}{ll}
4 & 5 \\
8 & 9
\end{array}\right)
$$
agrees with the usual method for computing the matrix product. (b) Use this method to compute the matrix products $(i)\left(\begin{array}{rr}-2 & 1 \\ 3 & 2\end{array}\right)\left(\begin{array}{rr}1 & -2 \\ 1 & 0\end{array}\right)$,
(ii)
$$
\left(\begin{array}{rrr}
1 & -2 & 0 \\
-3 & -1 & 2
\end{array}\right)\left(\begin{array}{rr}
2 & 5 \\
-3 & 0 \\
1 & -1
\end{array}\right)
$$
(iii) $\left(\begin{array}{rrr}3 & -1 & 1 \\ -1 & 2 & 1 \\ 1 & 1 & -5\end{array}\right)\left(\begin{array}{rrr}2 & 3 & 0 \\ 3 & -1 & 4 \\ 0 & 4 & 1\end{array}\right)$, and verify that you get the same answer as that obtained by the traditional method. (c) Explain why (1.13) is a special case of (1.14). (d) Prove that (1.14) gives the correct formula for the matrix product.

Key Concepts

Verification by Component Computation

Verifying matrix multiplication through specific examples by computing individual entries can illustrate that both the traditional approach and the sum of outer products method produce identical results. This technique demonstrates the internal consistency of the algebraic properties of matrices and provides a tangible check on more abstract theoretical formulations.

Equivalence of Multiplication Methods

Both the standard definition of matrix multiplication (row-by-column) and the alternative method (sum of outer products) yield the same result. This equivalence is due to the distributive property of matrix multiplication and the consistent way in which each entry in the resulting matrix is computed, reinforcing the robustness of the matrix product's definition.

Matrix Multiplication

Matrix multiplication is defined as the process of taking the dot product of rows of the first matrix with columns of the second matrix to produce a new matrix. This operation satisfies important properties such as associativity and distributivity, and it is fundamental in linear algebra for representing and solving systems of linear equations, transforming spaces, and many applications in applied mathematics and engineering.

Outer Product

The outer product is the multiplication of a column vector by a row vector, resulting in a matrix where each entry is the product of an element from the column and an element from the row. This construction yields a rank-one matrix and offers an alternative perspective on matrix multiplication, where the product of two matrices can be seen as the sum of multiple outer products.

Sum of Outer Products Representation

The sum of outer products representation expresses the matrix product as a sum of rank-one matrices, each formed by the outer product of a column of the first matrix and a corresponding row of the second matrix. This view is especially useful for both theoretical insights and computational strategies, as it underlines the structure of the resulting matrix through simpler components.

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