Talk:Dyadic product

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Is the tensor product of a row vector \mathbf{v} with a column vector \mathbf{u} still called a dyadic product? For example:

\mathbf{v} \otimes \mathbf{u} = \begin{bmatrix} v_1 & v_2 & v_3 \end{bmatrix} \otimes \begin{bmatrix} u_1 \\ u_2 \\ u_3 \end{bmatrix} = \begin{bmatrix} v_1u_1 & v_2u_1 & v_3u_1 \\ v_1u_2 & v_2u_2 & v_3u_2 \\ v_1u_3 & v_2u_3 & v_3u_3 \end{bmatrix}

The indices in the definitions would have to be swapped in that case. --RainerBlome 21:28, 27 August 2005 (UTC)

Technically, tensor products of vectors (and vectors themselves) are defined without a particular matrix structure in mind (i.e., "row" versus "column" is unimportant), so the question is one of representation only. The ordering of (column) x (row) is chosen to match the intuition of matrix multiplication, creating a square matrix that represents the dyadic product of the two original vectors. The only reason for using "column vector" and "row vector" is to make their matrix representations intuitive, so if we're going to change or generalize anything, we should emphasize that "row" and "column" are only important in representing dyadic products, not in defining them. Shiznick 06:02, 8 May 2007 (UTC)

Why/how is this any different than the Outer Product? Should it be merged? —Preceding unsigned comment added by 132.170.160.64 (talk) 22:56, 31 March 2008 (UTC)

An outer product of two vectors produces another vector, while the dyadic product produces a second-order tensor, a matrix. So, no merger in my opinion. Crowsnest (talk) 23:50, 31 March 2008 (UTC)