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Show that for every $A\in\mathbb{R}^{m\times n}$, $\textbf{v}\in\mathbb{R}^n$, $\textbf{w}\in\mathbb{R}^m$,

$$\left(A\textbf{v}\right)\cdot\textbf{w}=\textbf{v}\cdot\left(A^T\textbf{w}\right).$$

I know that I'm supposed to be using the summation convention $\textbf{v}=v_i\textbf{e}_i=\sum_{i=1}^{3}v_i\textbf{e}_i$ as well as the Levi-Civita symbol $\epsilon_{ijk}$ and the Kronecker delta $\delta_{ij}$.

Thanks!

PS: I reckon I could get the next part of the question if I could do the first part, but if you wanted to know, it's: Hence prove that if the matrix $Q\in\mathbb{R^{n\times n}}$ is orthogonal, that is if $Q^T=Q^{-1}$, then $$(Q\textbf{v})\cdot(Q\textbf{w})=\textbf{v}\cdot\textbf{w}$$ and $$\vert Q\textbf{v}\vert=\vert\textbf{v}\vert.$$

jamesh625
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    Why don't you just multiply out the formulae $\langle Av, w \rangle$, $\langle v, A^T w \rangle$ and see if they are the same? You can use the first result along with the fact that $Q^T Q = I$ to answer the second part. – copper.hat Jun 19 '13 at 05:18

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$$\begin{eqnarray*} \left(A\textbf{v}\right)\cdot\textbf{w} &=& (A{\bf v})_i w_i \\ &=& (A_{ij}v_j) w_i \\ &=& v_j (A_{ij} w_i) \\ &=& v_j ((A^T)_{ji}w_i) \\ &=& v_j (A^T{\bf w})_j \\ &=&\textbf{v}\cdot\left(A^T\textbf{w}\right) \end{eqnarray*}$$

user26872
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  • Ah great! Thanks oen. – jamesh625 Jun 19 '13 at 05:40
  • @user72786: Glad to help. Follow copper.hat's advice for the next part. – user26872 Jun 19 '13 at 05:44
  • Would this be the next part then?

    $$\left(Q\mathbf{v}\right)\cdot\left(Q\mathbf{w})\right) =\left(Q\mathbf{v}\right)i\left(Q\mathbf{w}\right)_i$$ $$=\left(Q{ij}v_j\right)\left(Q_{ij}w_j\right)$$ $$=Q_{ij}Q_{ij}\left(v_jw_j\right)$$ $$=Q_{ij}\left(Q^{T}\right)_{ji}\left(v_jw_j\right)$$ $$=QQ^{-1}\left(v_jw_j\right)$$ $$=\mathbf{v}\cdot\mathbf{w}.$$

     – jamesh625 Jun 19 '13 at 05:57
  • @user72786: We can now avoid components by using the above result: $(Q{\bf v})\cdot(Q{\bf w}) = {\bf v}\cdot(Q^TQ{\bf w}) = \ldots$. – user26872 Jun 19 '13 at 06:01
  • Oh great! :) Thanks again! – jamesh625 Jun 19 '13 at 06:10
  • @user72786: You're welcome. Here's how it looks in components - we need to be careful to use different dummy variables for different sums, $$\begin{eqnarray} \left(Q\textbf{v}\right)\cdot (Q \textbf{w}) &=& (Q{\bf v})i (Q{\bf w})_i \ &=& (Q{ij}v_j) (Q_{ik}w_k) \ &=& Q_{ij}Q_{ik} v_j w_k \ &=& (Q^T){ji}Q{ik} v_j w_k \ &=& (Q^T Q){jk} v_j w_k \ &=& I{jk} v_j w_k \ &=& \delta_{jk}v_j w_k \ &=& v_j w_j \ &=&\textbf{v}\cdot\textbf{w}. \end{eqnarray}$$ – user26872 Jun 19 '13 at 06:24
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\begin{align*} \langle Av, w \rangle &= (Av)^T w \\ &= v^T A^T w \\ &= \langle v, A^T w \rangle. \end{align*}

littleO
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