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Given $A=\left[\begin{array}{c} a_1 \\ a_2 \\ a_3 \end{array}\right]$ and $B=\left[\begin{array}{c} b_1 \\ b_2 \\ b_3 \end{array}\right]$ I would like to know the name the of the operator that produces the following matrix when I join $A$ and $B$: $\left[ \begin{matrix} (a_1,b_1) & (a_2,b_1) & (a_3,b_1) \\ (a_1,b_2) & (a_2,b_2) & (a_3,b_2) \\ (a_1,b_3) & (a_2,b_3) & (a_3,b_3) \end{matrix} \right]$.

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You want a matrix with entries in $\mathbb{R}^2$, right? –  Neal Jun 21 '12 at 20:17
    
Or is it an inner product? –  Calle Jun 21 '12 at 21:08
    
If $a_i,b_i$ are secretly row vectors, and $(,)$ means inner product, then you get the matrix $AB^T$. –  user31373 Jun 21 '12 at 21:35
    
I can express the matrix with canonical basis: $\left[ \begin{matrix} a_1\cdot e_1+b_1\cdot e_2 & a_2\cdot e_1+b_1\cdot e_2 & a_3\cdot e_1+b_1\cdot e_2 \\ a_1\cdot e_1+b_2\cdot e_2 & a_2\cdot e_1+b_2\cdot e_2 & a_3\cdot e_1+b_2\cdot e_2 \\ a_1\cdot e_1+b_3\cdot e_2 & a_2\cdot e_1+b_3\cdot e_2 & a_3\cdot e_1+b_3\cdot e_2 \end{matrix} \right]$. –  Paulo Fracasso Jun 21 '12 at 23:22
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I propose that we name this operator "The Fracasso embedding," in honor of its creator. –  Gerry Myerson Jul 5 '12 at 0:51
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