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Math Help - Algebra, Problems For Fun (25)

  1. #1
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    Algebra, Problems For Fun (25)

    This problem is a generalization of this simple fact that if \{a_1,a_2,a_3 \} \subset \mathbb{R}, then a_ia_j \geq 0 for some i \neq j:


    Let u_1, \cdots , u_{n+2} be any n+2 vectors in \mathbb{R}^n. Show that u_i \cdot u_j \geq 0 for some i \neq j.


    Suggestion:

    Spoiler:
    A clever induction over n.
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  2. #2
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    Thanks for this nice problem!

    By induction over n:
    - obvious if n=0 ( \mathbb{R}^0=\{0\}: the 0-dimensional vector space)
    - take n\geq 1 and assume this is true for n-1: among any (n+1)-uplet of vectors in \mathbb{R}^{n-1} (or in any (n-1)-dimensional Euclidean vector space, equivalently), there are (at least) two vectors that have nonnegative dot product. Let u_1,\ldots,u_{n+2}\in\mathbb{R}^n. If any vector is zero, the result is obvious, so we may assume u_{n+2}\neq 0. Denote by \pi the projection on the (n-1)-dimensional subspace u_{n+2}^\perp. By induction, there are 1\leq i\neq j\leq n+1 such that \pi(u_i)\cdot\pi(u_j)\geq 0. Then, consider u_i,u_j,u_{n+2}. I claim that one of u_i\cdot u_j, u_i\cdot u_{n+2} or u_j\cdot u_{n+2} is nonnegative. Indeed, if u_i\cdot u_{n+2}<0 and u_j\cdot u_{n+2}<0, then u_i\cdot u_j=\pi(u_i)\cdot\pi(u_j)+\frac{u_i\cdot u_{n+2}}{\|u_{n+2}\|}\times\frac{u_j\cdot u_{n+2}}{\|u_{n+2}\|}> 0 (because for any vector u, u=\pi(u)+\frac{u\cdot u_{n+2}}{\|u_{n+2}\|^2}u_{n+2}).
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