Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?


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 ?

share|cite|improve this question
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
up vote 1 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.