Let u , v ∈ R n . Prove that if ‖ u + t v ‖ ≥ ‖ u ‖for all t ∈ R , then u ⋅ v = 0 (vectors u and v are perpendicular).

Question

Let   u  ,  v  ∈            R        n  . Prove that if  ‖  u  +  t  v  ‖  ≥  ‖  u  ‖for all   t  ∈      R  , then   u  ⋅  v  =  0 (vectors u and v are perpendicular).

Hilfeform5c · Accepted Answer

‖  u      ‖          2        +      t          2        ‖  v      ‖          2        +  2  t  ⟨  u  ,  v  ⟩  ≥  ‖  u      ‖          2        for all t. This gives       t          2        ‖  v      ‖          2        +  2  t  ⟨  u  ,  v  ⟩  ≥  0. Take   t  &amp;gt;  0, divide by t and let   t  →  0 You get   ⟨  u  ,  v  ⟩  ≥  0. If you take limit through negative values you get the reverse inequality.

Alexander Lewis · Accepted Answer

|        |    u  +  t  v      |              |        2    ≥      |        |    u      |              |        2   , so   f  (  t  )  =      |        |    u  +  t  v      |              |        2   has global -- and therefore also local -- minimum at t=0, hence has       f    ′    (  0  )  =  2  u  ⋅  v  =  0

If norm(u+tv) >= norm(u) for all t, prove that u * v=0

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