Question

Question: Consider an $\mathbf{n}\mathbf{\times }\mathbf{m}$matrix A, a vector in ${\mathbf{R}}^{\mathbf{m}}$, and a vector $\overrightarrow{\mathit{w}}$in ${\mathbf{R}}^{\mathbf{m}}$. Show that $\left(A\overrightarrow{v}\right)\mathbf{.}\overrightarrow{\mathbf{w}}\mathbf{=}\overrightarrow{\mathbf{v}}\mathbf{.}\left(A^T\overrightarrow{w}\right)$.

Short answer

It is proved that $\left(\mathrm{A}\overrightarrow{\mathrm{v}}\right).\overrightarrow{\mathrm{w}}=\overrightarrow{\mathrm{v}}.\left({\mathrm{A}}^{\mathrm{T}}\overrightarrow{\mathrm{w}}\right)$.

Step-by-step solution

Consider the theorem

If$\overrightarrow{v}$ and $\overrightarrow{w}$are two (column) vectors in${\mathbf{R}}^{\mathbf{n}}$ , then $\begin{array}{cc}\stackrel{\overrightarrow{\mathbf{v}}\mathbf{.}\overrightarrow{\mathbf{w}}}{\mathbf{\uparrow }}& \stackrel{\mathbf{=}{\overrightarrow{\mathbf{v}}}^{\mathbf{T}}\overrightarrow{\mathbf{v}}}{\mathbf{\uparrow }}\\ \mathbf{}& \frac{\left(\mathrm{dot}\right)\left(\mathrm{matrix}\right)}{\mathbf{product}}\mathbf{}\end{array}$.

Application of the theorem

Observe that,

$$\begin{gathered} \left(A\overrightarrow{v}\right).\overrightarrow{w}={\left(A\overrightarrow{v}\right)}^T\overrightarrow{w}\left(theorem5.3.6\right) \\ =\left({\overrightarrow{v}}^T{A}^T\right)\overrightarrow{w\left(theorem5.3.9\left(c\right)\right)} \\ ={\overrightarrow{v}}^T\left(A^T\overrightarrow{w}\right)\left(Associativity\right) \\ =\overrightarrow{v}.\left(\left(A^T\overrightarrow{w}\right)\right) \end{gathered}$$According to the theorem.

Hence, $\left(A\overrightarrow{v}\right).\overrightarrow{w}=\overrightarrow{v}\left(A^T\overrightarrow{w}\right)$is proved.