Answer on Question #64854 – Math – Linear Algebra
Question
Find the orthogonal canonical reduction of the quadratic form
− x 2 + y 2 + z 2 + 2 x y − 2 x z + 2 y z - x ^ {2} + y ^ {2} + z ^ {2} + 2 x y - 2 x z + 2 y z − x 2 + y 2 + z 2 + 2 x y − 2 x z + 2 yz
Also, find its principal axes.
Solution
The matrix of the quadratic form:
A = ( − 1 1 − 1 1 1 1 − 1 1 1 ) A = \left( \begin{array}{ccc} - 1 & 1 & - 1\\ 1 & 1 & 1\\ -1 & 1 & 1 \end{array} \right) A = ⎝ ⎛ − 1 1 − 1 1 1 1 − 1 1 1 ⎠ ⎞
The characteristic equation:
∣ − 1 − λ 1 − 1 1 1 − λ 1 − 1 1 1 − λ ∣ = 0 \left| \begin{array}{ccc} - 1 - \lambda & 1 & - 1\\ 1 & 1 - \lambda & 1\\ -1 & 1 & 1 - \lambda \end{array} \right| = 0 ∣ ∣ − 1 − λ 1 − 1 1 1 − λ 1 − 1 1 1 − λ ∣ ∣ = 0 ( − 1 − λ ) ∣ 1 − λ 1 1 1 − λ ∣ − ∣ 1 1 − 1 1 − λ ∣ − ∣ 1 1 − λ − 1 1 ∣ = 0 (-1 - \lambda) \left| \begin{array}{cc} 1 - \lambda & 1 \\ 1 & 1 - \lambda \end{array} \right| - \left| \begin{array}{cc} 1 & 1 \\ -1 & 1 - \lambda \end{array} \right| - \left| \begin{array}{cc} 1 & 1 - \lambda \\ -1 & 1 \end{array} \right| = 0 ( − 1 − λ ) ∣ ∣ 1 − λ 1 1 1 − λ ∣ ∣ − ∣ ∣ 1 − 1 1 1 − λ ∣ ∣ − ∣ ∣ 1 − 1 1 − λ 1 ∣ ∣ = 0 ( − 1 − λ ) ( ( 1 − λ ) 2 − 1 ) − ( 1 − λ + 1 ) − ( 1 + 1 − λ ) = 0 (-1 - \lambda) ((1 - \lambda) ^ {2} - 1) - (1 - \lambda + 1) - (1 + 1 - \lambda) = 0 ( − 1 − λ ) (( 1 − λ ) 2 − 1 ) − ( 1 − λ + 1 ) − ( 1 + 1 − λ ) = 0 2 λ + 2 λ 2 − λ 2 − λ 3 + 2 λ − 4 = 0 2 \lambda + 2 \lambda^ {2} - \lambda^ {2} - \lambda^ {3} + 2 \lambda - 4 = 0 2 λ + 2 λ 2 − λ 2 − λ 3 + 2 λ − 4 = 0 λ 3 − λ 2 − 4 λ + 4 = 0 \lambda^ {3} - \lambda^ {2} - 4 \lambda + 4 = 0 λ 3 − λ 2 − 4 λ + 4 = 0 λ 2 ( λ − 1 ) − 4 ( λ − 1 ) = 0 \lambda^ {2} (\lambda - 1) - 4 (\lambda - 1) = 0 λ 2 ( λ − 1 ) − 4 ( λ − 1 ) = 0 ( λ 2 − 4 ) ( λ − 1 ) = 0 (\lambda^ {2} - 4) (\lambda - 1) = 0 ( λ 2 − 4 ) ( λ − 1 ) = 0 λ 1 = 1 ; λ 2 = − 2 ; λ 3 = 2 \lambda_ {1} = 1; \lambda_ {2} = - 2; \lambda_ {3} = 2 λ 1 = 1 ; λ 2 = − 2 ; λ 3 = 2
The orthogonal canonical reduction:
Q = λ 1 x ′ 2 + λ 2 y ′ 2 + λ 3 z ′ 2 Q = \lambda_ {1} x ^ {\prime 2} + \lambda_ {2} y ^ {\prime 2} + \lambda_ {3} z ^ {\prime 2} Q = λ 1 x ′2 + λ 2 y ′2 + λ 3 z ′2 Q = x ′ 2 − 2 y ′ 2 + 2 z ′ 2 Q = x ^ {\prime 2} - 2 y ^ {\prime 2} + 2 z ^ {\prime 2} Q = x ′2 − 2 y ′2 + 2 z ′2
For λ 1 = 1 \lambda_1 = 1 λ 1 = 1
( − 1 − 1 1 − 1 1 1 − 1 1 − 1 1 1 − 1 ) ( x y z ) = ( 0 0 0 ) \left( \begin{array}{c c c} - 1 - 1 & 1 & - 1 \\ 1 & 1 - 1 & 1 \\ - 1 & 1 & 1 - 1 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) ⎝ ⎛ − 1 − 1 1 − 1 1 1 − 1 1 − 1 1 1 − 1 ⎠ ⎞ ⎝ ⎛ x y z ⎠ ⎞ = ⎝ ⎛ 0 0 0 ⎠ ⎞ ( − 2 1 − 1 1 0 1 − 1 1 0 ) ( x y z ) = ( 0 0 0 ) \left( \begin{array}{c c c} - 2 & 1 & - 1 \\ 1 & 0 & 1 \\ - 1 & 1 & 0 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) ⎝ ⎛ − 2 1 − 1 1 0 1 − 1 1 0 ⎠ ⎞ ⎝ ⎛ x y z ⎠ ⎞ = ⎝ ⎛ 0 0 0 ⎠ ⎞ ( x y z ) = ( 1 1 − 1 ) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 1 \\ 1 \\ - 1 \end{array} \right) ⎝ ⎛ x y z ⎠ ⎞ = ⎝ ⎛ 1 1 − 1 ⎠ ⎞
The principal axis:
1 3 ( 1 1 − 1 ) \frac {1}{\sqrt {3}} \left( \begin{array}{c} 1 \\ 1 \\ - 1 \end{array} \right) 3 1 ⎝ ⎛ 1 1 − 1 ⎠ ⎞
For λ 2 = − 2 \lambda_{2} = -2 λ 2 = − 2
( − 1 + 2 1 − 1 1 1 + 2 1 − 1 1 1 + 2 ) ( x y z ) = ( 0 0 0 ) \left( \begin{array}{c c c} - 1 + 2 & 1 & - 1 \\ 1 & 1 + 2 & 1 \\ - 1 & 1 & 1 + 2 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) ⎝ ⎛ − 1 + 2 1 − 1 1 1 + 2 1 − 1 1 1 + 2 ⎠ ⎞ ⎝ ⎛ x y z ⎠ ⎞ = ⎝ ⎛ 0 0 0 ⎠ ⎞ ( 1 1 − 1 1 3 1 − 1 1 3 ) ( x y z ) = ( 0 0 0 ) \left( \begin{array}{c c c} 1 & 1 & - 1 \\ 1 & 3 & 1 \\ - 1 & 1 & 3 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) ⎝ ⎛ 1 1 − 1 1 3 1 − 1 1 3 ⎠ ⎞ ⎝ ⎛ x y z ⎠ ⎞ = ⎝ ⎛ 0 0 0 ⎠ ⎞ ( x y z ) = ( 2 − 1 1 ) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 2 \\ - 1 \\ 1 \end{array} \right) ⎝ ⎛ x y z ⎠ ⎞ = ⎝ ⎛ 2 − 1 1 ⎠ ⎞
The principal axis:
1 6 ( 2 − 1 1 ) \frac {1}{\sqrt {6}} \left( \begin{array}{c} 2 \\ - 1 \\ 1 \end{array} \right) 6 1 ⎝ ⎛ 2 − 1 1 ⎠ ⎞
For λ 3 = 2 \lambda_3 = 2 λ 3 = 2
( − 1 − 2 1 − 1 1 1 − 2 1 − 1 1 1 − 2 ) ( x y z ) = ( 0 0 0 ) \left( \begin{array}{c c c} - 1 - 2 & 1 & - 1 \\ 1 & 1 - 2 & 1 \\ - 1 & 1 & 1 - 2 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) ⎝ ⎛ − 1 − 2 1 − 1 1 1 − 2 1 − 1 1 1 − 2 ⎠ ⎞ ⎝ ⎛ x y z ⎠ ⎞ = ⎝ ⎛ 0 0 0 ⎠ ⎞ ( − 3 1 − 1 1 − 1 1 − 1 1 − 1 ) ( x y z ) = ( 0 0 0 ) \left( \begin{array}{c c c} - 3 & 1 & - 1 \\ 1 & - 1 & 1 \\ - 1 & 1 & - 1 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) ⎝ ⎛ − 3 1 − 1 1 − 1 1 − 1 1 − 1 ⎠ ⎞ ⎝ ⎛ x y z ⎠ ⎞ = ⎝ ⎛ 0 0 0 ⎠ ⎞ ( x y z ) = ( 0 1 1 ) . \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 0 \\ 1 \\ 1 \end{array} \right). ⎝ ⎛ x y z ⎠ ⎞ = ⎝ ⎛ 0 1 1 ⎠ ⎞ .
The principal axis:
1 2 ( 0 1 1 ) . \frac {1}{\sqrt {2}} \left( \begin{array}{c} 0 \\ 1 \\ 1 \end{array} \right). 2 1 ⎝ ⎛ 0 1 1 ⎠ ⎞ .
Answer: x ′ 2 − 2 y ′ 2 + 2 z ′ 2 ; 1 3 ( 1 1 ) , 1 6 ( 2 − 1 ) , 1 2 ( 0 1 ) . x^{\prime 2} - 2y^{\prime 2} + 2z^{\prime 2}; \quad \frac{1}{\sqrt{3}}\binom{1}{1}, \frac{1}{\sqrt{6}}\binom{2}{-1}, \frac{1}{\sqrt{2}}\binom{0}{1}. x ′2 − 2 y ′2 + 2 z ′2 ; 3 1 ( 1 1 ) , 6 1 ( − 1 2 ) , 2 1 ( 1 0 ) .
Answer provided by https://www.AssignmentExpert.com
#340153 on Dec 2023