Dot Product between a Dyad and a Vector

1. For calculating Dot Product between a Dyad and a Vector the Order of the Dyad Must be same as the Number of Elements in the Vector (i.e. Dimension of the Vector). The result of Dot Product between a Dyad and a Vector is another Vector.
2. Calculating Dot Product between a Dyad and a Vector is same as Multiplication of Dyad Matrix with the Vector. Calculating Dot Product between a Vector and a Dyad is same as Multiplication of Transpose of Dyad Matrix with the Vector OR Transpose of Multiplication of Transpose of Vector with the Dyad Matrix.
3. Given 3 Vectors \(\vec{A}\), \(\vec{B}\) and \(\vec{C}\), the following assosiative relations exist between Tensor Product (that results in a Dyad) and Dot Product
   
   \(\vec{C} \cdot (\vec{A} \otimes_{T} \vec{B})=\vec{C} \cdot \overleftrightarrow{AB} = (\vec{C} \cdot \vec{A}) \vec{B}\)
   
   \((\vec{A} \otimes_{T} \vec{B}) \cdot \vec{C} =\overleftrightarrow{AB} \cdot \vec{C}= \vec{A} (\vec{B} \cdot \vec{C})\)
   
   
4. The Dot Product between a Dyad and a Vector is Non-Commutative, that is for Any Non Identity Dyad/Dyadic \(\overleftrightarrow{AB}\) and Vector \(\vec{C}\)
   
   \(\overleftrightarrow{AB} \cdot \vec{C} \neq \vec{C} \cdot \overleftrightarrow{AB}\)
5. Following examples demonstrates the calculation Dot Product between a Dyad and a Vector and assosiative relations that exist between Tensor Product (that results in a Dyad) and Dot Product for Vectors \(\vec{A}\), \(\vec{B}\) and \(\vec{C}\) as given below
   
   \(\vec{A}=\begin{bmatrix}12\\6\\4\end{bmatrix}\hspace{.5cm}\vec{B}=\begin{bmatrix}-4\\1\\-3\end{bmatrix}\hspace{.5cm}\vec{C}=\begin{bmatrix}5\\2\\-4\end{bmatrix}\)
   
   \(\overleftrightarrow{AB}=\vec{A} \otimes_{T} \vec{B}=\begin{bmatrix}12\\6\\4\end{bmatrix} \otimes_{T} \begin{bmatrix}-4 \\ 1 \\ -3\end{bmatrix}=\vec{A} \otimes \vec{B}^{T}=\begin{bmatrix}12\\6\\4\end{bmatrix} \otimes \begin{bmatrix}-4 & 1 & -3\end{bmatrix}=\begin{bmatrix}-48 & 12 & -36 \\ -24 & 6 & -18 \\ -16 & 4 & -12\end{bmatrix}\)
   
   Now,
   
   \(\vec{C} \cdot \overleftrightarrow{AB}= {[{\vec{C}}^T\overleftrightarrow{AB}]}^T={\overleftrightarrow{AB}}^T\vec{C}=\begin{bmatrix}-48 & -24 & -16 \\ 12 & 6 & 4 \\ -36 & -18 & -12\end{bmatrix}\begin{bmatrix}5\\2\\-4\end{bmatrix}=\begin{bmatrix}-224\\56\\-168\end{bmatrix}\)   ...(1)
   
   Also,
   
   \((\vec{C} \cdot \vec{A})\vec{B}=(\begin{bmatrix}5\\2\\-4\end{bmatrix} \cdot \begin{bmatrix}12\\6\\4\end{bmatrix})\begin{bmatrix}-4\\1\\-3\end{bmatrix}=56 \begin{bmatrix}-4\\1\\-3\end{bmatrix}=\begin{bmatrix}-224\\56\\-168\end{bmatrix}\)   ...(2)
   
   From equations (1) and (2) we get
   
   \(\vec{C} \cdot (\vec{A} \otimes_{T} \vec{B})=\vec{C} \cdot \overleftrightarrow{AB} = (\vec{C} \cdot \vec{A}) \vec{B}\)
   
   Similarly,
   
   \(\overleftrightarrow{AB} \cdot \vec{C} = {\overleftrightarrow{AB}}\vec{C}=\begin{bmatrix}-48 & 12 & -36 \\ -24 & 6 & -18 \\ -16 & 4 & -12\end{bmatrix}\begin{bmatrix}5\\2\\-4\end{bmatrix}=\begin{bmatrix}-72\\-36\\-24\end{bmatrix}\)   ...(3)
   
   Also,
   
   \(\vec{A} (\vec{B}\cdot \vec{C})=\begin{bmatrix}12\\6\\4\end{bmatrix}(\begin{bmatrix}-4\\1\\-3\end{bmatrix}\cdot \begin{bmatrix}5\\2\\-4\end{bmatrix})=-6 \begin{bmatrix}12\\6\\4\end{bmatrix}=\begin{bmatrix}-72\\-36\\-124\end{bmatrix}\)   ...(4)
   
   From equations (3) and (4) we get
   
   \((\vec{A} \otimes_{T} \vec{B}) \cdot \vec{C} =\overleftrightarrow{AB} \cdot \vec{C}= \vec{A} (\vec{B} \cdot \vec{C})\)