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Given two pairs of vector (U, V) and $(U_1, V_1)$ where U, V, $U_1$, $V_1$ have the same length K, $U_1$ and $V_1$ are constant vectors. Is it possible to say that minimizing $\|U-U_1 \|^2$ + $\|V-V_1 \|^2$ is equivalent to minimize $(U^TV-U^T_1V_1)^2$ ? If it's possible, how to prove it ?

Thanks.

PS: Some comments below point out that we cannot have the way from second expression to the first expression. Can we prove minimizing the first expression => minimize the second one ?

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I think they are not equivalent. For the first expression to be zero we need $U=U_1$ and $V=V_1$. But the second one could become zero at many other possibilities, i.e. only the dot products have to match; meaning the angle between $U$ and $V$ should be same as angle between $U_1$ and $V_1$. –  Maesumi Mar 1 '13 at 3:31

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Your original expression is equal to $4K^2 -2U^TU_1-2V^TV_1$ by expanding the inner products. This is easier to optimise (perhaps).

No they are not equivalent. You will have the same angle between $U$ and $V$ as between $U_1$ and $V_1$ in your reformulation, but they may not be good approximations for what you are trying to achieve.

Counterexample for your reformulation:

$U=[1,0],\,V=[1,1]$ and $U_1=[-5,3],\,V_1=[1,2]$. You can readily check the dot products of the pairs of vectors are both $1$, but the vectors are by no means the optimum for your original problem.

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which expression ? –  mr noname Mar 1 '13 at 3:40
    
In this case, can I prove the one way that minimizing the first expression will also minimize the second one ? –  mr noname Mar 1 '13 at 3:47
    
@mrnoname The optimal solution of the first expression is $U=U_1,\,V=V_1$. Clearly, this will make the second expression zero, which is it's optimum. In general, for an approximate optimal solution $\tilde{U}$ and $\tilde{V}$, I don't have a pen and paper with me to verify the optimality of the second expression. –  Daryl Mar 1 '13 at 3:52
    
I did try like this. $\Delta U = U_1-U$, $\Delta V = V_1-V$. $(U^T_1V_1-U^TV)^2=(U^T_1\Delta V + V^T_1\Delta U - \Delta U^T \Delta V)^2$. Then I am stuck. Because $\|\Delta U\|^2+\|\Delta V\|^2 \rightarrow 0$ is not strong enough to say $(U^T_1V_1-U^TV)^2=(U^T_1\Delta V + V^T_1\Delta U - \Delta U^T \Delta V)^2 \rightarrow 0$ with $U_1$ and $V_1$ are constant vectors. –  mr noname Mar 1 '13 at 4:02

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