Equivalence of geometric and algebraic definition of dot product

Question

Reading the wikipedia dot product page I am confused by the last equation:

$A \cdot B = \sum B_i(A \cdot e_i) = \sum A_iB_i$

Specifically how is the $A \cdot e_i$ part derived? How is it substituted into that equation? Also why is $A \cdot e_i = A_i$ How is this equation related to the previous equation: $A \cdot B = A_B \|B\|$ ?

presumably, $e_i$ is the vector $(0,\dots,0,1,0,\dots,0)$ of all zeros except for a 1 at the $i$th entry. So $A\cdot e_i=A_i$. — Gerry Myerson, Dec 26 '13 at 23:26
Ok that makes sense. How do they get from A ⋅ B to $\sum B_i(A⋅e_i)$ ? — Bryan Glazer, Dec 27 '13 at 22:48
From $B=\sum B_ie_i$, they go $A\cdot B=A\cdot\sum B_ie_i=\sum A\cdot(B_ie_i)=\sum B_i(A\cdot e_i)$. — Gerry Myerson, Dec 27 '13 at 23:01
Ah that seems so obvious in retrospect. Thanks. If you add this as an answer I will accept it. — Bryan Glazer, Dec 28 '13 at 23:00

score 1 · Accepted Answer · answered Dec 29 '13 at 01:19

1

At author's suggestion, my comment, as an answer:

From $B=\sum B_ie_i$, it goes $$A\cdot B=A\cdot\sum B_ie_i=\sum A\cdot(B_ie_i)=\sum B_i(A\cdot e_i)$$

answered Dec 29 '13 at 01:19

Gerry Myerson

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Equivalence of geometric and algebraic definition of dot product

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