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Combinatus

- 42

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## 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.

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