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In my book we defined $$(\mathbf{a}\otimes\mathbf{b})_{ij} = a_ib_j$$ then the book goes on and says the outer product is distributive. Then it does the following $$\mathbf{a}\otimes \mathbf{b} = (a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k})\otimes(b_1\mathbf{i}+b_2\mathbf{j}+b_3\mathbf{k})= a_1b_1\mathbf{i}\otimes\mathbf{i}+a_1b_2\mathbf{i}\otimes\mathbf{j}+a_1b_3\mathbf{i}\otimes\mathbf{k}+..$$ and continues saying that this equals $$..=\sum_{i=1}^3a_ib_j\hat{\mathbf{e}}_i\otimes\hat{\mathbf{e}}_j = a_ib_j\hat{\mathbf{e}}_i\otimes\hat{\mathbf{e}}_j$$

I'm at the beginning in the topic but.. surely the penultimate equality is wrong? Shouldn't it be this? $$\sum_{j=1}^3\sum_{i=1}^3a_ib_j\hat{\mathbf{e}}_i\otimes\hat{\mathbf{e}}_j$$ indeed if you expand the outer product you get $9$ terms, whereas in the above given by the book you get only $3$ terms, and how do you find the $j$ values? The only way I could think it is correct is by using somehow the Einstein notation convention, but I don't see how, here $a_i$ and $b_j$ have "hanging indexes", not "dummy" ones.

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    Looks like a typo, or an unwise abbreviation. There are 9 terms (although 3 are automatically 0, when i=j) – 2017-01-31
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    How do you know that it is zero when $i =j$? – 2017-01-31
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    It must be a typo. The expression that you wrote is definitely the right one. Also, none of the 9 terms are in general zero. – 2017-01-31
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    @NafizIshtiaqueOrnob thank you! – 2017-01-31
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    When $i=j$ we have $e_i\times e_j=e_i\times e_i=0.$ We have $ v\times v=0$ for all $v. $ – 2017-02-01
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    @user254665 mmm I thought outer product and cross product were two different things.. – 2017-02-01

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