Matrix multiplication distributivity

Question

Suppose we have matrices $A, B, C$ of dimensions $m \times n, m\times n, n \times l$ respectively. How can we prove $(A+B)\circ C = A\circ C + B \circ C$ (using the summation notation method?)

@LePressentiment Please don't edit someone's question just because the notation isn't perfectly standard. Some people use $\circ$ to emphasise that the multiplication is not usual multiplication of real numbers and it's perfectly acceptable to use such notation. — Dan Rust, Aug 13 '13 at 14:26

score 1 · Answer 1 · answered Mar 21 '13 at 23:07

1

Hint: Look at the linear maps represented by $A$, $B$ and $C$.

answered Mar 21 '13 at 23:07

azimut

22,696

I'm mostly having trouble reconciling the summation notation. – user67928 Mar 21 '13 at 23:07

score 0 · Accepted Answer · answered Mar 21 '13 at 23:24

Tedious way...coefficientwise: $$ ((A+B)C)_{i,j}=\sum_{k=1}^n(A+B)_{i,k}C_{k,j}=\sum_{k=1}^n(A_{i,k}+B_{i,k})C_{k,j} $$ $$ =\sum_{k=1}^nA_{i,k}C_{k,j}+B_{i,k}C_{k,j}=\sum_{k=1}^nA_{i,k}C_{k,j}+\sum_{k=1}^nB_{i,k}C_{k,j} $$ $$ (AC)_{i,j}+(BC)_{i,j}=(AC+BC)_{i,j} $$ for all $1\leq i\leq n$ and all $1\leq j\leq l$. So $(A+B)C=AC+BC$.

score 0 · Answer 3 · answered Mar 22 '13 at 00:11

0

Let $L = (A+B)C, R = AC + BC$.

With the Einstein summation convention, you have

$$L^i_j = (A+B)^i_k C^k_j = (A^i_k+B^i_k) C^k_j = A^i_k C^k_j + B^i_k C^k_j = R^i_j$$

answered Mar 22 '13 at 00:11

copper.hat

172,524

Matrix multiplication distributivity

3 Answers3