# Matrix multiplication distributivity

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?)

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@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. –  Daniel Rust Aug 13 '13 at 14:26

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 $(A+B)C=AC+BC$.

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

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

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

With the Einstein summation convention, you have

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

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