Quadratic form (maxima and minima problem)

[Combinatus]

Homework Statement

Determine the largest value of $$x_{1}^2 + x_{2}^2 + x_{3}^2$$ when $$x_{1}^2 + 2x_{2}^2 + 2x_{3}^2 + 2x_{1}x_{2} + 2x_{1}x_{3} + 2x_{2}x_{3} = 1$$

Homework Equations

Not sufficiently relevant to produce the expected answer.

The Attempt at a Solution

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

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

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

$$x_{1}^2 + x_{2}^2 + x_{3}^2 = (x_{1}' - x_{2}' - x_{3}')^2 + x_{2}'^2 + x_{3}'^2 = x_{1}'^2 + 2x_{2}'^2 + 2x_{3}'^2 - 2x_{1}'x_{2}' - 2x_{1}'x_{3}' + 2x_{2}'x_{3}'$$ (1)

when $$x_{1}'^2 + x_{2}'^2 + x_{3}'^2 = 1$$.

Rewriting (1), we get: $$x_{1}'^2 + 2x_{2}'^2 + 2x_{3}'^2 - 2x_{1}'x_{2}' - 2x_{1}'x_{3}' + 2x_{2}'x_{3}' =<br /> <br /> \begin{pmatrix}x_{1}' & x_{2}' & x_{3}' \end{pmatrix}<br /> <br /> \begin{pmatrix}<br /> <br /> 1 & -1 & -1 \\<br /> -1 & 2 & 1 \\<br /> -1 & 1 & 2 \end{pmatrix}<br /> <br /> \begin{pmatrix}x_{1}' \\ x_{2}' \\ x_{3}' \end{pmatrix} = X'^tAX' = X''^tDX'' = \lambda_{1} x''_{1}^2 + \lambda_{2} x''_{2}^2 + \lambda_{3} x''_{3}^2<br />$$

by using the spectral theorem, the relation $$X' = TX''$$ 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 $$\begin{vmatrix} 1-\lambda & -1 & -1\\-1 & 2-\lambda & 1\\-1 & 1 & 2-\lambda \end{vmatrix} = 0 \iff (\lambda - 1)(\lambda - (2-\sqrt{3}))(\lambda-(2+\sqrt{3})) = 0$$

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

Since $$|X'|^2 = x_{1}'^2 + x_{2}'^2 + x_{3}'^2 = x_{1}''^2 + x_{2}''^2 + x_{3}''^2 = 1$$ (all bases involved are orthonormal), it follows that the largest value sought is equal to the largest eigenvalue, $$2+\sqrt{3}$$, and the smallest value sought is equal to the smallest eigenvalue, $$2-\sqrt{3}$$.Turns out that the key to the problem doesn't agree. It states that the largest value is $$1/(2-\sqrt{3})$$ and the smallest is $$1/(2+\sqrt{3})$$. 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.

 

[Gib Z]

The key to the problem should have rationalized the denominator of its answers =)

 

[Combinatus]

UGH. http://intp.se/whoco5.gif Note to self: FFFFUUUUUUUUUUU.

Thank you.