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Homework Help: Quadratic form (maxima and minima problem)

  1. Mar 3, 2010 #1
    1. The problem statement, all variables and given/known data

    Determine the largest value of [tex]x_{1}^2 + x_{2}^2 + x_{3}^2[/tex] when [tex]x_{1}^2 + 2x_{2}^2 + 2x_{3}^2 + 2x_{1}x_{2} + 2x_{1}x_{3} + 2x_{2}x_{3} = 1[/tex]


    2. Relevant equations

    Not sufficiently relevant to produce the expected answer.


    3. The attempt at a solution

    Completing the square, we get [tex]x_{1}^2 + 2x_{2}^2 + 2x_{3}^2 + 2x_{1}x_{2} + 2x_{1}x_{3} + 2x_{2}x_{3} = (x_1 + x_2 + x_3)^2 + x_{2}^2 + x_{3}^2 = 1[/tex].

    Assigning [tex]x_{1}' = x_{1} + x_{2} + x_{3}, x_{2}' = x_{2}, x_{3}' = x_{3}[/tex] as a coordinate change (and checking that the determinant of the transformation matrix is non-zero to ensure its validity), we equivalently get [tex]x_{1} = x_{1}' - x_{2}' - x_{3}', x_{2} = x_{2}', x_{3} = x_{3}'[/tex].

    Consequently, the problem can be formulated as determining the largest and smallest value of

    [tex]x_{1}^2 + x_{2}^2 + x_{3}^2 = (x_{1}' - x_{2}' - x_{3}')^2 + x_{2}'^2 + x_{3}'^2 = x_{1}'^2 + 2x_{2}'^2 + 2x_{3}'^2 - 2x_{1}'x_{2}' - 2x_{1}'x_{3}' + 2x_{2}'x_{3}'[/tex] (1)

    when [tex]x_{1}'^2 + x_{2}'^2 + x_{3}'^2 = 1[/tex].

    Rewriting (1), we get: [tex]x_{1}'^2 + 2x_{2}'^2 + 2x_{3}'^2 - 2x_{1}'x_{2}' - 2x_{1}'x_{3}' + 2x_{2}'x_{3}' =

    \begin{pmatrix}x_{1}' & x_{2}' & x_{3}' \end{pmatrix}

    \begin{pmatrix}

    1 & -1 & -1 \\
    -1 & 2 & 1 \\
    -1 & 1 & 2 \end{pmatrix}

    \begin{pmatrix}x_{1}' \\ x_{2}' \\ x_{3}' \end{pmatrix} = X'^tAX' = X''^tDX'' = \lambda_{1} x''_{1}^2 + \lambda_{2} x''_{2}^2 + \lambda_{3} x''_{3}^2

    [/tex]

    by using the spectral theorem, the relation [tex]X' = TX''[/tex] between the bases and the orthogonality of transformation matrices between orthonormal bases. (T and D are obviously the transformation matrix between the bases and the diagonalized matrix in the orthonormal eigenvector base, respectively.)

    Solving the characteristic equation [tex]\begin{vmatrix} 1-\lambda & -1 & -1\\-1 & 2-\lambda & 1\\-1 & 1 & 2-\lambda \end{vmatrix} = 0 \iff (\lambda - 1)(\lambda - (2-\sqrt{3}))(\lambda-(2+\sqrt{3})) = 0[/tex]

    we get that [tex]X''^t D X'' = x''_{1}^2 + (2-\sqrt{3}) x''_{2}^2 + (2+\sqrt{3}) x''_{3}^2[/tex]

    Since [tex]|X'|^2 = x_{1}'^2 + x_{2}'^2 + x_{3}'^2 = x_{1}''^2 + x_{2}''^2 + x_{3}''^2 = 1[/tex] (all bases involved are orthonormal), it follows that the largest value sought is equal to the largest eigenvalue, [tex]2+\sqrt{3}[/tex], and the smallest value sought is equal to the smallest eigenvalue, [tex]2-\sqrt{3}[/tex].


    Turns out that the key to the problem doesn't agree. It states that the largest value is [tex]1/(2-\sqrt{3})[/tex] and the smallest is [tex]1/(2+\sqrt{3})[/tex]. If I knew where I purportedly managed to mess up, you would not be reading this sentence. Some assistance, or perhaps a suggestion for a different approach, will be appreciated.
     
    Last edited: Mar 3, 2010
  2. jcsd
  3. Mar 3, 2010 #2

    Gib Z

    User Avatar
    Homework Helper

    The key to the problem should have rationalized the denominator of its answers =)
     
  4. Mar 3, 2010 #3
    UGH. http://intp.se/whoco5.gif [Broken] Note to self: FFFFUUUUUUUUUUU.

    Thank you.
     
    Last edited by a moderator: May 4, 2017
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