Dyadics

Dyadic, outer, and tensor products A dyad is a tensor of order two and rank one, and is the dyadic product of two vectors (complex vectors in general), whereas a dyadic is a general tensor of order two (which may be full rank or not). There are several equivalent terms and notations for this product: • the dyadic product of two vectors \mathbf{a} and \mathbf{b} is denoted by \mathbf{a}\mathbf{b} (juxtaposed; no symbols, multiplication signs, crosses, dots, etc.) • the outer product of two column vectors \mathbf{a} and \mathbf{b} is denoted and defined as \mathbf{a} \otimes \mathbf{b} or \mathbf{a}\mathbf{b}^\mathsf{T}, where \mathsf{T} means transpose, • the tensor product of two vectors \mathbf{a} and \mathbf{b} is denoted \mathbf{a} \otimes \mathbf{b}, In the dyadic context they all have the same definition and meaning, and are used synonymously, although the tensor product is an instance of the more general and abstract use of the term. Three-dimensional Euclidean space To illustrate the equivalent usage, consider three-dimensional Euclidean space, letting: : \begin{align} \mathbf{a} &= a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} \\ \mathbf{b} &= b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k} \end{align} be two vectors where i, j, k (also denoted e1, e2, e3) are the standard basis vectors in this vector space (see also Cartesian coordinates). Then the dyadic product of a and b can be represented as a sum: : \begin{align} \mathbf{ab} =\qquad &a_1 b_1 \mathbf{ii} + a_1 b_2 \mathbf{ij} + a_1 b_3 \mathbf{ik} \\ {}+{} &a_2 b_1 \mathbf{ji} + a_2 b_2 \mathbf{jj} + a_2 b_3 \mathbf{jk} \\ {}+{} &a_3 b_1 \mathbf{ki} + a_3 b_2 \mathbf{kj} + a_3 b_3 \mathbf{kk} \end{align} or by extension from row and column vectors, a 3×3 matrix (also the result of the outer product or tensor product of a and b): : \mathbf{a b} \equiv \mathbf{a}\otimes\mathbf{b} \equiv \mathbf{a b}^\mathsf{T} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}\begin{pmatrix} b_1 & b_2 & b_3 \end{pmatrix} = \begin{pmatrix} a_1 b_1 & a_1 b_2 & a_1 b_3 \\ a_2 b_1 & a_2 b_2 & a_2 b_3 \\ a_3 b_1 & a_3 b_2 & a_3 b_3 \end{pmatrix}. Just as the standard basis (and unit) vectors i, j, k, have the representations: : \begin{align} \mathbf{i} &= \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},& \mathbf{j} &= \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix},& \mathbf{k} &= \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \end{align} (which can be transposed), the standard basis (and unit) dyads have the representation: : \begin{align} \mathbf{ii} &= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, & \mathbf{ij} &= \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, & \mathbf{ik} &= \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \\ \mathbf{ji} &= \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, & \mathbf{jj} &= \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}, & \mathbf{jk} &= \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \\ \mathbf{ki} &= \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix}, & \mathbf{kj} &= \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}, & \mathbf{kk} &= \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \end{align} For a simple numerical example in the standard basis: : \begin{align} \mathbf{A} &= 2\mathbf{ij} + \frac{\sqrt{3}}{2}\mathbf{ji} - 8\pi\mathbf{jk} + \frac{2\sqrt{2}}{3}\mathbf{kk} \\[2pt] &= 2 \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} + \frac{\sqrt{3}}{2}\begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} - 8\pi \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} + \frac{2\sqrt{2}}{3}\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \\[2pt] &= \begin{pmatrix} 0 & 2 & 0 \\ \frac{\sqrt{3}}{2} & 0 & -8\pi \\ 0 & 0 & \frac{2\sqrt{2}}{3} \end{pmatrix} \end{align} N-dimensional Euclidean space If the Euclidean space is N-dimensional, and : \begin{align} \mathbf{a} &= \sum_{i=1}^N a_i\mathbf{e}_i = a_1 \mathbf{e}_1 + a_2 \mathbf{e}_2 + {\ldots} + a_N \mathbf{e}_N \\ \mathbf{b} &= \sum_{j=1}^N b_j\mathbf{e}_j = b_1 \mathbf{e}_1 + b_2 \mathbf{e}_2 + \ldots + b_N \mathbf{e}_N \end{align} where ei and ej are the standard basis vectors in N-dimensions (the index i on ei selects a specific vector, not a component of the vector as in ai), then in algebraic form their dyadic product is: : \mathbf{ab} = \sum_{j=1}^N \sum_{i=1}^N a_i b_j \mathbf{e}_i \mathbf{e}_j. This is known as the nonion form of the dyad. Their outer/tensor product in matrix form is: : \mathbf{ab} = \mathbf{ab}^\mathsf{T} = \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_N \end{pmatrix}\begin{pmatrix} b_1 & b_2 & \cdots & b_N \end{pmatrix} = \begin{pmatrix} a_1 b_1 & a_1 b_2 & \cdots & a_1 b_N \\ a_2 b_1 & a_2 b_2 & \cdots & a_2 b_N \\ \vdots & \vdots & \ddots & \vdots \\ a_N b_1 & a_N b_2 & \cdots & a_N b_N \end{pmatrix}. A dyadic polynomial A, otherwise known as a dyadic, is formed from multiple vectors ai and bj: : \mathbf{A} = \sum_i\mathbf{a}_i\mathbf{b}_i = \mathbf{a}_1\mathbf{b}_1 + \mathbf{a}_2\mathbf{b}_2 + \mathbf{a}_3\mathbf{b}_3 + \ldots A dyadic which cannot be reduced to a sum of less than N dyads is said to be complete. In this case, the forming vectors are non-coplanar, see Chen (1983). Classification The following table classifies dyadics: : Identities The following identities are a direct consequence of the definition of the tensor product: {{ordered list : (\alpha\mathbf{a}) \mathbf{b} = \mathbf{a} (\alpha\mathbf{b}) = \alpha (\mathbf{a}\mathbf{b}) for any scalar \alpha. : \begin{align} \mathbf{a} (\mathbf{b} + \mathbf{c}) &= \mathbf{a}\mathbf{b} + \mathbf{a}\mathbf{c} \\ (\mathbf{a} + \mathbf{b}) \mathbf{c} &= \mathbf{a}\mathbf{c} + \mathbf{b}\mathbf{c} \end{align} }} == Dyadic algebra ==

Product of dyadic and vector There are four operations defined on a vector and dyadic, constructed from the products defined on vectors. : Product of dyadic and dyadic There are five operations for a dyadic to another dyadic. Let a, b, c, d be real vectors. Then: : } \left(\mathbf{ab}\right) \mathbin{{}_\times^{\,\centerdot}} \left(\mathbf{cd}\right) = \left(\mathbf{a}\times\mathbf{c}\right)\left(\mathbf{b}\cdot\mathbf{d}\right) \left(\mathbf{ab}\right) \mathbin{{}_\times^\times} \left(\mathbf{cd}\right) = \left(\mathbf{a}\times\mathbf{c}\right)\left(\mathbf{b}\times\mathbf{d}\right) Letting : \mathbf{A} = \sum_i \mathbf{a}_i\mathbf{b}_i,\quad \mathbf{B} = \sum_j \mathbf{c}_j\mathbf{d}_j be two general dyadics, we have: : } \mathbf{A} \mathbin{{}_\times^{\,\centerdot}} \mathbf{B} = \sum_{i,j} \left(\mathbf{a}_i\times\mathbf{c}_j\right)\left(\mathbf{b}_i\cdot\mathbf{d}_j\right) \mathbf{A} \mathbin{{}_\times^\times} \mathbf{B} = \sum_{i,j} \left(\mathbf{a}_i\times \mathbf{c}_j\right)\left(\mathbf{b}_i\times \mathbf{d}_j\right) Double–dot product The first definition of the double–dot product is the Frobenius inner product, : \begin{align} \operatorname{tr}\left(\mathbf{A}\mathbf{B}^\mathsf{T}\right) &=\sum_{i,j} \operatorname{tr}\left(\mathbf{a}_i \mathbf{b}_i^\mathsf{T} \mathbf{d}_j \mathbf{c}_j^\mathsf{T}\right) \\ &=\sum_{i,j} \operatorname{tr}\left(\mathbf{c}_j^\mathsf{T} \mathbf{a}_i \mathbf{b}_i^\mathsf{T} \mathbf{d}_j\right) \\ &=\sum_{i,j} (\mathbf{a}_i\cdot\mathbf{c}_j)(\mathbf{b}_i\cdot\mathbf{d}_j) \\ &=\mathbf{A} \mathbin{{}_\centerdot^\centerdot} \mathbf{B} \end{align} Furthermore, since, : \begin{align} \mathbf{A}^\mathsf{T} &=\sum_{i,j} \left(\mathbf{a}_i\mathbf{b}_j^\mathsf{T}\right)^\mathsf{T} \\ &=\sum_{i,j} \mathbf{b}_i\mathbf{a}_j^\mathsf{T} \end{align} we get that, : \mathbf{A} \mathbin{{}_\centerdot^\centerdot} \mathbf{B} = \mathbf{A} \mathbin{\underline{{}_\centerdot^\centerdot}} \mathbf{B}^\mathsf{T} so the second possible definition of the double-dot product is just the first with an additional transposition on the second dyadic. For these reasons, the first definition of the double-dot product is preferred, though some authors still use the second. Double–cross product We can see that, for any dyad formed from two vectors a and b, its double cross product is zero. : \left(\mathbf{ab}\right) \mathbin{{}_\times^\times} \left(\mathbf{ab}\right) = \left(\mathbf{a}\times\mathbf{a}\right)\left(\mathbf{b}\times\mathbf{b}\right) = 0 However, by definition, a dyadic double-cross product on itself will generally be non-zero. For example, a dyadic A composed of six different vectors : \mathbf{A} = \sum_{i=1}^3 \mathbf{a}_i\mathbf{b}_i has a non-zero self-double-cross product of : \mathbf{A} \mathbin{{}_\times^\times} \mathbf{A} = 2\left[ \left(\mathbf{a}_1 \times \mathbf{a}_2\right)\left(\mathbf{b}_1 \times \mathbf{b}_2\right) + \left(\mathbf{a}_2 \times \mathbf{a}_3\right)\left(\mathbf{b}_2 \times \mathbf{b}_3\right) + \left(\mathbf{a}_3 \times \mathbf{a}_1\right)\left(\mathbf{b}_3 \times \mathbf{b}_1\right) \right] Tensor contraction The spur or expansion factor arises from the formal expansion of the dyadic in a coordinate basis by replacing each dyadic product by a dot product of vectors: : \begin{align} |\mathbf{A}| =\qquad &A_{11} \mathbf{i}\cdot\mathbf{i} + A_{12} \mathbf{i}\cdot\mathbf{j} + A_{13} \mathbf{i}\cdot\mathbf{k} \\ {}+{} &A_{21} \mathbf{j}\cdot\mathbf{i} + A_{22} \mathbf{j}\cdot\mathbf{j} + A_{23} \mathbf{j}\cdot\mathbf{k} \\ {}+{} &A_{31} \mathbf{k}\cdot\mathbf{i} + A_{32} \mathbf{k}\cdot\mathbf{j} + A_{33} \mathbf{k}\cdot\mathbf{k} \\[6pt] =\qquad &A_{11} + A_{22} + A_{33} \end{align} in index notation this is the contraction of indices on the dyadic: : |\mathbf{A}| = \sum_i A_i{}^i In three dimensions only, the rotation factor arises by replacing every dyadic product by a cross product : \begin{align} \langle\mathbf{A}\rangle =\qquad &A_{11} \mathbf{i}\times\mathbf{i} + A_{12} \mathbf{i}\times\mathbf{j} + A_{13} \mathbf{i}\times\mathbf{k} \\ {}+{} &A_{21} \mathbf{j}\times\mathbf{i} + A_{22} \mathbf{j}\times\mathbf{j} + A_{23} \mathbf{j}\times\mathbf{k}\\ {}+{} &A_{31} \mathbf{k}\times\mathbf{i} + A_{32} \mathbf{k}\times\mathbf{j} + A_{33} \mathbf{k}\times\mathbf{k} \\[6pt] =\qquad &A_{12} \mathbf{k} - A_{13} \mathbf{j} - A_{21} \mathbf{k} \\ {}+{} &A_{23} \mathbf{i} + A_{31} \mathbf{j} - A_{32} \mathbf{i} \\[6pt] =\qquad &\left(A_{23} - A_{32}\right)\mathbf{i} + \left(A_{31} - A_{13}\right)\mathbf{j} + \left(A_{12} - A_{21}\right)\mathbf{k}\\ \end{align} In index notation this is the contraction of A with the Levi-Civita tensor : \langle\mathbf{A}\rangle = \sum_{jk}{\epsilon_i}^{jk}A_{jk}. == Unit dyadic ==

There exists a unit dyadic, denoted by I, such that, for any vector a, : \mathbf{I}\cdot\mathbf{a}=\mathbf{a}\cdot\mathbf{I}= \mathbf{a} Given a basis of 3 vectors a, b and c, with reciprocal basis \hat{{\mathbf{a}}}, \hat{\mathbf{b}}, \hat{\mathbf{c}}, the unit dyadic is expressed by : \mathbf{I} = \mathbf{a}\hat{\mathbf{a}} + \mathbf{b}\hat{\mathbf{b}} + \mathbf{c}\hat{\mathbf{c}} In the standard basis (for definitions of i, j, k see in the above section ''''), : \mathbf{I} = \mathbf{ii} + \mathbf{jj} + \mathbf{kk} Explicitly, the dot product to the right of the unit dyadic is : \begin{align} \mathbf{I} \cdot \mathbf{a} & = (\mathbf{i}\mathbf{i} + \mathbf{j}\mathbf{j} + \mathbf{k}\mathbf{k})\cdot \mathbf{a} \\ & = \mathbf{i}(\mathbf{i} \cdot \mathbf{a}) + \mathbf{j}(\mathbf{j} \cdot \mathbf{a}) + \mathbf{k} (\mathbf{k} \cdot \mathbf{a}) \\ & = \mathbf{i} a_x + \mathbf{j} a_y + \mathbf{k} a_z \\ & = \mathbf{a} \end{align} and to the left : \begin{align} \mathbf{a} \cdot \mathbf{I} & = \mathbf{a} \cdot (\mathbf{i}\mathbf{i} + \mathbf{j}\mathbf{j} + \mathbf{k}\mathbf{k})\\ & = (\mathbf{a}\cdot \mathbf{i})\mathbf{i} + (\mathbf{a}\cdot \mathbf{j})\mathbf{j} + (\mathbf{a}\cdot \mathbf{k})\mathbf{k} \\ & = a_x \mathbf{i} + a_y \mathbf{j} + a_z \mathbf{k} \\ & = \mathbf{a} \end{align} The corresponding matrix is : \mathbf{I}=\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{pmatrix} This can be put on more careful foundations (explaining what the logical content of "juxtaposing notation" could possibly mean) using the language of tensor products. If V is a finite-dimensional vector space, a dyadic tensor on V is an elementary tensor in the tensor product of V with its dual space. The tensor product of V and its dual space is isomorphic to the space of linear maps from V to V: a dyadic tensor vf is simply the linear map sending any w in V to f(w)v. When V is Euclidean n-space, we can use the inner product to identify the dual space with V itself, making a dyadic tensor an elementary tensor product of two vectors in Euclidean space. In this sense, the unit dyadic ij is the function from 3-space to itself sending a1i + a2j + a3k to a2i, and jj sends this sum to a2j. Now it is revealed in what (precise) sense ii + jj + kk is the identity: it sends a1i + a2j + a3k to itself because its effect is to sum each unit vector in the standard basis scaled by the coefficient of the vector in that basis. Properties of unit dyadics : \begin{align} \left(\mathbf{a}\times\mathbf{I}\right)\cdot\left(\mathbf{b}\times\mathbf{I}\right) &= \mathbf{ba} - \left(\mathbf{a}\cdot\mathbf{b}\right)\mathbf{I} \\ \mathbf{I} {}_\times^{\,\centerdot} \left(\mathbf{ab}\right) &= \mathbf{b}\times\mathbf{a} \\ \mathbf{I} {}_\times^\times \mathbf{A} &= (\mathbf{A} {}_{\,\centerdot}^{\,\centerdot} \mathbf{I})\mathbf{I} - \mathbf{A}^\mathsf{T} \\ \mathbf{I} {}_{\,\centerdot}^{\,\centerdot} \left(\mathbf{ab}\right) &= \left(\mathbf{I}\cdot\mathbf{a}\right)\cdot\mathbf{b} = \mathbf{a}\cdot\mathbf{b} = \mathrm{tr}\left(\mathbf{ab}\right) \end{align} where "tr" denotes the trace. == Examples ==

Vector projection and rejection A nonzero vector a can always be split into two perpendicular components, one parallel (‖) to the direction of a unit vector n, and one perpendicular (⊥) to it; : \mathbf{a} = \mathbf{a}_\parallel + \mathbf{a}_\perp The parallel component is found by vector projection, which is equivalent to the dot product of a with the dyadic nn, : \mathbf{a}_\parallel = \mathbf{n}(\mathbf{n}\cdot\mathbf{a}) = (\mathbf{nn})\cdot\mathbf{a} and the perpendicular component is found from vector rejection, which is equivalent to the dot product of a with the dyadic , : \mathbf{a}_\perp = \mathbf{a} - \mathbf{n}(\mathbf{n}\cdot\mathbf{a}) = (\mathbf{I} - \mathbf{nn})\cdot\mathbf{a} Rotation dyadic 2d rotations The dyadic : \mathbf{J} = \mathbf{ji} - \mathbf{ij} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} is a 90° anticlockwise rotation operator in 2d. It can be left-dotted with a vector r = xi + yj to produce the vector, : (\mathbf{j i} - \mathbf{i j}) \cdot (x \mathbf{i} + y \mathbf{j}) = x \mathbf{j i} \cdot \mathbf{i} - x \mathbf{i j} \cdot \mathbf{i} + y \mathbf{j i} \cdot \mathbf{j} - y \mathbf{i j} \cdot \mathbf{j} = -y \mathbf{i} + x \mathbf{j}, in summary : \mathbf{J}\cdot\mathbf{r} = \mathbf{r}_\mathrm{rot} or in matrix notation : \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}= \begin{pmatrix} -y \\ x \end{pmatrix}. For any angle θ, the 2d rotation dyadic for a rotation anti-clockwise in the plane is : \mathbf{R} = \mathbf{I}\cos\theta + \mathbf{J}\sin\theta = (\mathbf{ii}+\mathbf{jj})\cos\theta + (\mathbf{ji}-\mathbf{ij})\sin\theta = \begin{pmatrix} \cos\theta &-\sin\theta \\ \sin\theta &\;\cos\theta \end{pmatrix} where I and J are as above, and the rotation of any 2d vector a = axi + ayj is : \mathbf{a}_\mathrm{rot} = \mathbf{R}\cdot\mathbf{a} 3d rotations A general 3d rotation of a vector a, about an axis in the direction of a unit vector ω and anticlockwise through angle θ, can be performed using Rodrigues' rotation formula in the dyadic form : \mathbf{a}_\mathrm{rot} = \mathbf{R} \cdot \mathbf{a} \,, where the rotation dyadic is : \mathbf{R} = \mathbf{I} \cos\theta + \boldsymbol{\Omega} \sin\theta + \boldsymbol{\omega\omega} (1 - \cos\theta) \,, and the Cartesian entries of ω also form those of the dyadic : \boldsymbol{\Omega} = \omega_x( \mathbf{kj} - \mathbf{jk} ) + \omega_y( \mathbf{ik} - \mathbf{ki} ) + \omega_z( \mathbf{ji} - \mathbf{ij} ) \,, The effect of Ω on a is the cross product : \boldsymbol{\Omega} \cdot \mathbf{a} = \boldsymbol{\omega} \times \mathbf{a} which is the dyadic form the cross product matrix with a column vector. Lorentz transformation In special relativity, the Lorentz boost with speed v in the direction of a unit vector n can be expressed as : t' = \gamma\left(t - \frac{v\mathbf{n}\cdot\mathbf{r}}{c^2} \right) : \mathbf{r}' = [\mathbf{I} + (\gamma-1) \mathbf{nn}]\cdot \mathbf{r} - \gamma v \mathbf{n}t where : \gamma=\frac{1}{\sqrt{1-\dfrac{v^2}{c^2}}} is the Lorentz factor. == Related terms ==

Some authors generalize from the term dyadic to related terms triadic, tetradic and polyadic. == See also ==