- #1
Combinatus
- 42
- 1
Homework Statement
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]
Homework Equations
Not sufficiently relevant to produce the expected answer.
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: