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}&#039;^2 + 2x_{2}&#039;^2 + 2x_{3}&#039;^2 - 2x_{1}&#039;x_{2}&#039; - 2x_{1}&#039;x_{3}&#039; + 2x_{2}&#039;x_{3}&#039; =<br /> <br /> \begin{pmatrix}x_{1}&#039; &amp; x_{2}&#039; &amp; x_{3}&#039; \end{pmatrix}<br /> <br /> \begin{pmatrix}<br /> <br /> 1 &amp; -1 &amp; -1 \\<br /> -1 &amp; 2 &amp; 1 \\<br /> -1 &amp; 1 &amp; 2 \end{pmatrix}<br /> <br /> \begin{pmatrix}x_{1}&#039; \\ x_{2}&#039; \\ x_{3}&#039; \end{pmatrix} = X&#039;^tAX&#039; = X&#039;&#039;^tDX&#039;&#039; = \lambda_{1} x&#039;&#039;_{1}^2 + \lambda_{2} x&#039;&#039;_{2}^2 + \lambda_{3} x&#039;&#039;_{3}^2<br /> <br />

by using the spectral theorem, the relation X&#039; = TX&#039;&#039; 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 &amp; -1 &amp; -1\\-1 &amp; 2-\lambda &amp; 1\\-1 &amp; 1 &amp; 2-\lambda \end{vmatrix} = 0 \iff (\lambda - 1)(\lambda - (2-\sqrt{3}))(\lambda-(2+\sqrt{3})) = 0

we get that X&#039;&#039;^t D X&#039;&#039; = x&#039;&#039;_{1}^2 + (2-\sqrt{3}) x&#039;&#039;_{2}^2 + (2+\sqrt{3}) x&#039;&#039;_{3}^2

Since |X&#039;|^2 = x_{1}&#039;^2 + x_{2}&#039;^2 + x_{3}&#039;^2 = x_{1}&#039;&#039;^2 + x_{2}&#039;&#039;^2 + x_{3}&#039;&#039;^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.