# what is the right notation for this operation on two vectors?

I am trying to find the right notation for this operation on two vectors of size N and M. I do not believe it is the dot product, because there is no sum for the multiplication of each element. Instead, the result is a vector that is of size N * M:

v1 = [a,b,c] v2 = [d,e]

result: [ad,ae,bd,be,cd,ce]

What is the notation for the operation I just performed between v1 and v2?

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## 1 Answer

In a certain ordering of the basis, that's a tensor product, denoted $v_1 \otimes v_2$.

The tensor product $V \otimes W$ of vector spaces has a formal definition in terms of a universal property, but one way you can concretely construct it: if you're given a basis $\{e_1, \dots, e_n\}$ of $V$ and a basis $\{f_1, \dots, f_m\}$ of $W$, then $V \otimes W$ is an $mn$-dimensional vector space whose basis you denote by $\{e_i \otimes f_j\}$ for $i = 1, \dots, n$ and $j = 1, \dots, m$.

Given any vectors $v \in V$ and $w \in W$, you can form a corresponding vector $v \otimes w$ in the tensor product space as follows: Write $v = \sum_{i=1}^n c_i e_i$ and $w = \sum_{j=1}^m d_j f_j$; then define the vector $v \otimes w$ to be $$\sum_{i=1}^n \sum_{j=1}^m c_i d_j (e_i \otimes f_j).$$ This assignment is bilinear, meaning that if you fix $v$, then $v \otimes w$ varies linearly with $w$, and if you fix $w$ then $v \otimes w$ varies linearly with $v$.

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hmmm, i'm wondering if it is the cartesian product? – gnychis Feb 28 '12 at 2:31
Yes, the tensor product is related to the cartesian product; if I is the index set for a basis of V, and J is the index set for a basis of W, then the cartesian product of I and J is the index set for a basis of the tensor product of V and W. – Dave Gaebler Mar 7 '12 at 22:42