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Let $\mathbf{v}=(v_1,\cdots,v_n)^T, \mathbf{w}=(w_1,\cdots,w_n)^T, \mathbf{a}=(a_1,\cdots,a_n)^T \in\mathbb{R}^n$, and let $$ A = (\mathbf{w}\cdot\mathbf{a})(\mathbf{v}\cdot\mathbf{a})=\left(\sum_{i=1}^{n}w_ia_i\right)\left(\sum_{j=1}^{n}v_ja_j\right) $$ Is there any way of simplifying this quantity? Into one sum, maybe?

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2 Answers 2

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You can write it equivalently as either $(w^Ta)(v^Ta)=(w^Ta)^T(v^Ta)=a^Twv^Ta=a^T(wv^T)a$, or

$(v^Ta)(w^Ta)= ...=a^T(vw^T)a$.

Therefore, $(w^Ta)(v^Ta) = a^TBa$ is a quadratic form with the symmetric matrix $B=\frac{1}{2}(w v^T + v w^T )$.

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Well, you could make it the difference of two squares of sums:

$$\left(\sum_{i=1}^n \left(\dfrac{w_i + v_i}{2}\right) a_i \right)^2 - \left(\sum_{i=1}^n \left(\dfrac{w_i - v_i}{2}\right) a_i \right)^2$$

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    $\begingroup$ I didn't understanding, how this simplies the quantity? $\endgroup$
    – DiegoMath
    Jun 24, 2014 at 15:01
  • $\begingroup$ Yes, me neither! Thanks, anyway! $\endgroup$ Jun 24, 2014 at 15:23
  • $\begingroup$ I just wonder if there is any "clever" way of simplifying $A$ that I am unable to see now. Probably there is none! But if anyone can see it, let it be known! $\endgroup$ Jun 24, 2014 at 15:39

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