The matrix of this quadratic form is
⎝⎛3−10−12−10−13⎠⎞
We have to find its eigenvalues and eigenvectors.
det∣∣3−x−10−12−x−10−13−x∣∣=(3−x)2(2−x)−2(3−x)
=(3−x)((3−x)(2−x)−2)=(3−x)(x2−5x+4)=(3−x)(x−1)(x−4)
Therefore, the eigenvalues are 1, 3 and 4.
1) Consider the eigenvalue 1:
⎝⎛3−1−10−12−1−10−13−1⎠⎞⎝⎛xyz⎠⎞=⎝⎛000⎠⎞
⎝⎛2−10−11−10−12⎠⎞⎝⎛xyz⎠⎞=⎝⎛000⎠⎞
2x−y=0, 2z−y=0, therefore, the eigenvector is
⎝⎛xyz⎠⎞=⎝⎛212⎠⎞
Its norm is 22+12+22=3. Therefore, the normalized eigenvector, corressponding to the eigenvalue 1, is ⎝⎛xyz⎠⎞=⎝⎛2/31/32/3⎠⎞.
2) Consider the eigenvalue 3:
⎝⎛3−3−10−12−3−10−13−3⎠⎞⎝⎛xyz⎠⎞=⎝⎛000⎠⎞
⎝⎛0−10−1−1−10−10⎠⎞⎝⎛xyz⎠⎞=⎝⎛000⎠⎞
−y=0, −x−y−z=0, therefore, the eigenvector is
⎝⎛xyz⎠⎞=⎝⎛10−1⎠⎞
Its norm is 12+02+(−1)2=2. Therefore, the normalized eigenvector, corressponding to the eigenvalue 3, is ⎝⎛1/20−1/2⎠⎞.
3) Consider the eigenvalue 4:
⎝⎛3−4−10−12−4−10−13−4⎠⎞⎝⎛xyz⎠⎞=⎝⎛000⎠⎞
⎝⎛−1−10−1−2−10−1−1⎠⎞⎝⎛xyz⎠⎞=⎝⎛000⎠⎞
−x−y=0, −y−z=0, therefore, the eigenvector is
⎝⎛xyz⎠⎞=⎝⎛1−11⎠⎞
Its norm is 12+(−1)2+12=3. Therefore, the normalized eigenvector, corressponding to the eigenvalue 3, is ⎝⎛1/3−1/31/3⎠⎞.
By the construction we have
⎝⎛3−10−12−10−13⎠⎞⎝⎛2/31/32/31/20−1/21/3−1/31/3⎠⎞=
=⎝⎛2/31/32/31/20−1/21/3−1/31/3⎠⎞⎝⎛100030004⎠⎞
and
3x2+2y2+3z2−2xy−2yz=(xyz)⎝⎛3−10−12−10−13⎠⎞⎝⎛xyz⎠⎞=
(xyz)⎝⎛2/31/32/31/20−1/21/3−1/31/3⎠⎞⎝⎛100030004⎠⎞⎝⎛2/31/32/31/20−1/21/3−1/31/3⎠⎞−1⎝⎛xyz⎠⎞
But the matrix
⎝⎛2/31/32/31/20−1/21/3−1/31/3⎠⎞ is orthogonal, since it is composed with the orthonormal eigenvectors of the symmetric matrice. therefore
⎝⎛2/31/32/31/20−1/21/3−1/31/3⎠⎞−1=⎝⎛2/31/32/31/20−1/21/3−1/31/3⎠⎞T
=⎝⎛2/31/32/31/20−1/21/3−1/31/3⎠⎞
Put
⎝⎛2/31/32/31/20−1/21/3−1/31/3⎠⎞⎝⎛xyz⎠⎞=⎝⎛XYZ⎠⎞
then
3x2+2y2+3z2−2xy−2yz=(XYZ)⎝⎛100030004⎠⎞⎝⎛XYZ⎠⎞=X2+3Y2+4Z2
And we reduce the given quadratic form to the canonical form by orthogonal transformation.
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