Home > Algebra, Higher Algebra > SEEMOUS 2011 Problem 3

SEEMOUS 2011 Problem 3


Given vectors \overline{a},\overline{b},\overline{c} \in \Bbb{R}^n, show that
(\| \overline{a}\|\langle \overline{b},\overline{c}\rangle)^2+(\| \overline{b}\|\langle \overline{a},\overline{c}\rangle)^2\leq \|\overline{a}\|\|\overline{b}\|(\|\overline{a}\|\|\overline{b}\|+|\langle \overline{a},\overline{b}\rangle |)\|\overline{c}\|^2, with the usual notations.
SEEMOUS 2010 Problem 3

Solution: First, reduce the problem to \Bbb{R}^3. How we can do that? Consider an orthonormal basis for which the first three vectors span the vector space spanned by a,b,c. An orthonormal transformation preserves the norm and the inner product, therefore, it would be enough to prove the inequality in the new basis, a basis in which a,b,c have only the first three components non-zero. Therefore, if we solve the problem for a,b,c \in \Bbb{R}^3 we are done. But in \Bbb{R}^3 we can reduce the problem to a geometric one. First, note that changing the sign of some of the vectors a,b,c does not change anything in the inequality. If we denote \alpha=\angle (b,c),\ \beta=\angle (a,c),\ \gamma=\angle (a,b). Arrange the signs such that \alpha ,\beta \in [0,\pi/2] and drawing a trihedron angle OABC with supports of its sides determined by a,b,c the projection of C on the plane OAB would be inside the angle \angle OAB.

It is easy to prove that if x,y,z satisfy x+y=z then \cos^2 x + \cos^2 y+\cos^2 z-2\cos x\cos y\cos z=1. The inequality we want to prove is \cos^2 \alpha +\cos^2 \beta\leq 1 + |\cos \gamma|. Note that if one of \alpha,\beta is \pi/2 then we are done.

Project C on the plane OAB in point P, and denote \alpha_1=\angle POB,\beta_1=\angle POA. Project P on OA,OB in A_1,B_1. By expressing cosine in a right triangle, we get \cos^2 \alpha=OB_1^2/OC^2\leq OB_1^2/OP^2=\cos^2 \alpha_1. Similar to this we get \cos^2 \beta \leq \cos^2 \beta_1. Use the given identity for \alpha_1+\beta_1=\gamma and get \cos^2 \alpha_1+\cos^2\beta_1 =1+2 \cos \alpha_1 \cos \beta_1 \cos \gamma -\cos^2 \gamma . Let’s prove that

$latex 1+2\cos\alpha_1\cos\beta_1\cos\gamma-\cos^2\gamma$ \leq 1 +\cos \gamma . Of course, if \cos \gamma=0 we are done. Else, the inequality is equivalent to 2 \cos \alpha_1 \cos \beta_1 \leq 1+\cos \gamma=\cos(\alpha_1+\beta_1)=1+\cos\alpha_1 \cos \beta_1 -\sin \alpha_1 \sin \beta_1 . Finally, this last inequality is equivalent to \cos(\alpha_1 -\beta_1) \leq 1 , which is true.

This solves the problem for n \geq 3. For n=1 the inequality is trivial, and for n=2 we do something similar to the arguments above, but we are no longer in space, and in the plane, the problem is even easier.

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  1. nicu beli
    March 9, 2011 at 11:11 am

    Eu tot geometric ma gandisem. Si tot ca tine am presupus ca \alpha,\beta\in [0,\pi /2]. Trebuie sa aratam ca |\cos\gamma |\geq\cos^2\alpha +\cos^2\beta -1=\frac 12(\cos 2\alpha +\cos 2\beta )=\cos (\alpha+\beta )\cos (\alpha -\beta ). Avem |\alpha -\beta |\leq\pi /2 deci \cos (\alpha -\beta )\geq 0 deci putem presupune ca \cos (\alpha+\beta ),\cos (\alpha -\beta )>0. Insa \alpha,\beta,\gamma sunt unghirile triedrului 0abc (care poate fi si degenerat) deci avem 0\leq\gamma\leq\alpha +\beta\leq\pi. Rezulta ca 0<\cos (\alpha +\beta )\leq\cos\gamma, care, impreuna cu 0<\cos (\alpha -\beta )\leq 1 implica \cos (\alpha+\beta )\cos (\alpha -\beta )\leq\cos\gamma.

    • March 9, 2011 at 12:36 pm

      E adevarat ca la problema asta s-au dat mai multe solutii distincte? Asa am auzit.

  2. dani
    March 12, 2011 at 12:31 am

    O solutie foarte rapida ar fi sa scriem determinantul Gramm pentru 3 vectori oarecare a,b,c avand dimensiunea n. Stim ca are determinantul mai mare sau egal decat 0 si se obtine imediat concluzia.

    (Enghish: A fast solution would be to write the Gramm determinant for the three vectors.)

  3. zoe
    March 16, 2011 at 10:58 am

    Ce stiti despre studentul care a fost descalificat pentru ca a dat solutiile din lista scurta?

    • March 16, 2011 at 11:49 am

      Nu am auzit despre acest incident. Daca intr-adevar au fost scurgeri de informatii de la profesori catre studenti, acest fapt este foarte trist.

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