Answer on Question #42303 – Math – Linear Algebra:
a) Let a quadratic form have the expression x 2 + y 2 + 2 z 2 + 2 x y + 3 x z x^{2} + y^{2} + 2z^{2} + 2xy + 3xz x 2 + y 2 + 2 z 2 + 2 x y + 3 x z B 1 = { ( 1 , 0 , 0 ) , ( 0 , 1 , 0 ) , ( 0 , 0 , 1 ) } B_{1} = \{(1,0,0),(0,1,0),(0,0,1)\} B 1 = {( 1 , 0 , 0 ) , ( 0 , 1 , 0 ) , ( 0 , 0 , 1 )} B 2 = { ( 1 , 1 , 1 ) , ( 0 , 1 , 0 ) , ( 0 , 1 , 1 ) } B_{2} = \{(1,1,1),(0,1,0),(0,1,1)\} B 2 = {( 1 , 1 , 1 ) , ( 0 , 1 , 0 ) , ( 0 , 1 , 1 )}
b) Consider the quadratic form Q ( x , y , z ) = 2 x 2 − 4 x y + y 2 + 4 x z + 3 z 2 Q(x,y,z) = 2x^{2} - 4xy + y^{2} + 4xz + 3z^{2} Q ( x , y , z ) = 2 x 2 − 4 x y + y 2 + 4 x z + 3 z 2
i) Find a symmetric matrix A A A Q = X T A X Q = X^T AX Q = X T A X
ii) Find the orthogonal canonical reduction of the quadratic form.
iii) Find the principal axes of the form.
iv) Find the rank and signature of the form.
Solution.
a)
{ u = x + y + z v = y w = y + z ⇒ { x = u − y − z y = v z = w − y ⇒ { x = u − w y = v z = w − v ; \left\{ \begin{array}{l} u = x + y + z \\ v = y \\ w = y + z \end{array} \right. \Rightarrow \left\{ \begin{array}{c} x = u - y - z \\ y = v \\ z = w - y \end{array} \right. \Rightarrow \left\{ \begin{array}{c} x = u - w \\ y = v \\ z = w - v \end{array} ; \right. ⎩ ⎨ ⎧ u = x + y + z v = y w = y + z ⇒ ⎩ ⎨ ⎧ x = u − y − z y = v z = w − y ⇒ ⎩ ⎨ ⎧ x = u − w y = v z = w − v ; Q ( x , y , z ) = x 2 + y 2 + 2 z 2 + 2 x y + 3 x z = = ( u − w ) 2 + v 2 + 2 ( w − v ) 2 + 2 v ( u − w ) + 3 ( u − w ) ( w − v ) = = u 2 − 2 u w + w 2 + v 2 + 2 ( w 2 − 2 v w + v 2 ) + 2 ( u v − v w ) + + 3 ( u w − u v + v w − w 2 ) = u 2 + 3 v 2 − u v + u w − 3 v w . \begin{array}{l} Q (x, y, z) = x ^ {2} + y ^ {2} + 2 z ^ {2} + 2 x y + 3 x z = \\ = (u - w) ^ {2} + v ^ {2} + 2 (w - v) ^ {2} + 2 v (u - w) + 3 (u - w) (w - v) = \\ = u ^ {2} - 2 u w + w ^ {2} + v ^ {2} + 2 \left(w ^ {2} - 2 v w + v ^ {2}\right) + 2 (u v - v w) + \\ + 3 (u w - u v + v w - w ^ {2}) = u ^ {2} + 3 v ^ {2} - u v + u w - 3 v w. \\ \end{array} Q ( x , y , z ) = x 2 + y 2 + 2 z 2 + 2 x y + 3 x z = = ( u − w ) 2 + v 2 + 2 ( w − v ) 2 + 2 v ( u − w ) + 3 ( u − w ) ( w − v ) = = u 2 − 2 u w + w 2 + v 2 + 2 ( w 2 − 2 v w + v 2 ) + 2 ( uv − v w ) + + 3 ( u w − uv + v w − w 2 ) = u 2 + 3 v 2 − uv + u w − 3 v w .
Hence:
Q ( u , v , w ) = u 2 + 3 v 2 − u v + u w − 3 v w . Q (u, v, w) = u ^ {2} + 3 v ^ {2} - u v + u w - 3 v w. Q ( u , v , w ) = u 2 + 3 v 2 − uv + u w − 3 v w .
b)
i) Q ( x , y , z ) = 2 x 2 − 4 x y + y 2 + 4 x z + 3 z 2 = 2 ( x 2 + 2 x ( z − y ) ) + y 2 + 3 z 2 = 2 ( x + z − y ) 2 − 2 ( z − y ) 2 + y 2 + 3 z 2 = 2 ( x + z − y ) 2 − y 2 + z 2 + 4 y z = 2 ( x − y + z ) 2 + ( z + 2 y ) 2 − 4 y 2 − y 2 = 2 ( x − y + z ) 2 − 5 y 2 + ( 2 y + z ) 2 ; Q(x,y,z) = 2x^{2} - 4xy + y^{2} + 4xz + 3z^{2} = 2\big(x^{2} + 2x(z - y)\big) + y^{2} + 3z^{2} = 2(x + z - y)^{2} - 2(z - y)^{2} + y^{2} + 3z^{2} = 2(x + z - y)^{2} - y^{2} + z^{2} + 4yz = 2(x - y + z)^{2} + (z + 2y)^{2} - 4y^{2} - y^{2} = 2(x - y + z)^{2} - 5y^{2} + (2y + z)^{2}; Q ( x , y , z ) = 2 x 2 − 4 x y + y 2 + 4 x z + 3 z 2 = 2 ( x 2 + 2 x ( z − y ) ) + y 2 + 3 z 2 = 2 ( x + z − y ) 2 − 2 ( z − y ) 2 + y 2 + 3 z 2 = 2 ( x + z − y ) 2 − y 2 + z 2 + 4 yz = 2 ( x − y + z ) 2 + ( z + 2 y ) 2 − 4 y 2 − y 2 = 2 ( x − y + z ) 2 − 5 y 2 + ( 2 y + z ) 2 ;
( u v w ) = ( x − y + z y 2 y + z ) = ( 1 − 1 1 0 1 0 0 2 1 ) ( x y z ) ; Q = ( u v w ) T ( 2 0 0 0 − 5 0 0 0 1 ) ( u v w ) = = ( x y z ) T ( 1 − 1 1 0 1 0 0 2 1 ) T ( 2 0 0 0 − 5 0 0 0 1 ) ( 1 − 1 1 0 1 0 0 2 1 ) ( x y z ) = = ( x y z ) T ( 1 0 0 − 1 1 2 1 0 1 ) ( 2 0 0 0 − 5 0 0 0 1 ) ( 1 − 1 1 0 1 0 0 2 1 ) ( x y z ) = = ( x y z ) T ( 2 0 0 − 2 − 5 2 2 0 1 ) ( 1 − 1 1 0 1 0 0 2 1 ) ( x y z ) = ( x y z ) T ( 2 − 2 2 − 2 1 0 2 0 3 ) ( x y z ) ; \begin{array}{l} \left( \begin{array}{c} u \\ v \\ w \end{array} \right) = \left( \begin{array}{c} x - y + z \\ y \\ 2 y + z \end{array} \right) = \left( \begin{array}{c c c} 1 & - 1 & 1 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right); \\ Q = \left( \begin{array}{c} u \\ v \\ w \end{array} \right) ^ {T} \left( \begin{array}{c c c} 2 & 0 & 0 \\ 0 & - 5 & 0 \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{c} u \\ v \\ w \end{array} \right) = \\ = \left( \begin{array}{c} x \\ y \\ z \end{array} \right) ^ {T} \left( \begin{array}{c c c} 1 & - 1 & 1 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{array} \right) ^ {T} \left( \begin{array}{c c c} 2 & 0 & 0 \\ 0 & - 5 & 0 \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{c c c} 1 & - 1 & 1 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \\ = \left( \begin{array}{c} x \\ y \\ z \end{array} \right) ^ {T} \left( \begin{array}{c c c} 1 & 0 & 0 \\ - 1 & 1 & 2 \\ 1 & 0 & 1 \end{array} \right) \left( \begin{array}{c c c} 2 & 0 & 0 \\ 0 & - 5 & 0 \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{c c c} 1 & - 1 & 1 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \\ = \left( \begin{array}{c} x \\ y \\ z \end{array} \right) ^ {T} \left( \begin{array}{c c c} 2 & 0 & 0 \\ - 2 & - 5 & 2 \\ 2 & 0 & 1 \end{array} \right) \left( \begin{array}{c c c} 1 & - 1 & 1 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} x \\ y \\ z \end{array} \right) ^ {T} \left( \begin{array}{c c c} 2 & - 2 & 2 \\ - 2 & 1 & 0 \\ 2 & 0 & 3 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right); \\ \end{array} ⎝ ⎛ u v w ⎠ ⎞ = ⎝ ⎛ x − y + z y 2 y + z ⎠ ⎞ = ⎝ ⎛ 1 0 0 − 1 1 2 1 0 1 ⎠ ⎞ ⎝ ⎛ x y z ⎠ ⎞ ; Q = ⎝ ⎛ u v w ⎠ ⎞ T ⎝ ⎛ 2 0 0 0 − 5 0 0 0 1 ⎠ ⎞ ⎝ ⎛ u v w ⎠ ⎞ = = ⎝ ⎛ x y z ⎠ ⎞ T ⎝ ⎛ 1 0 0 − 1 1 2 1 0 1 ⎠ ⎞ T ⎝ ⎛ 2 0 0 0 − 5 0 0 0 1 ⎠ ⎞ ⎝ ⎛ 1 0 0 − 1 1 2 1 0 1 ⎠ ⎞ ⎝ ⎛ x y z ⎠ ⎞ = = ⎝ ⎛ x y z ⎠ ⎞ T ⎝ ⎛ 1 − 1 1 0 1 0 0 2 1 ⎠ ⎞ ⎝ ⎛ 2 0 0 0 − 5 0 0 0 1 ⎠ ⎞ ⎝ ⎛ 1 0 0 − 1 1 2 1 0 1 ⎠ ⎞ ⎝ ⎛ x y z ⎠ ⎞ = = ⎝ ⎛ x y z ⎠ ⎞ T ⎝ ⎛ 2 − 2 2 0 − 5 0 0 2 1 ⎠ ⎞ ⎝ ⎛ 1 0 0 − 1 1 2 1 0 1 ⎠ ⎞ ⎝ ⎛ x y z ⎠ ⎞ = ⎝ ⎛ x y z ⎠ ⎞ T ⎝ ⎛ 2 − 2 2 − 2 1 0 2 0 3 ⎠ ⎞ ⎝ ⎛ x y z ⎠ ⎞ ;
Hence:
A = ( 2 − 2 2 − 2 1 0 2 0 3 ) . A = \left( \begin{array}{c c c} 2 & - 2 & 2 \\ - 2 & 1 & 0 \\ 2 & 0 & 3 \end{array} \right). A = ⎝ ⎛ 2 − 2 2 − 2 1 0 2 0 3 ⎠ ⎞ .
ii)
( u v w ) = ( x − y + z y 2 y + z ) ⇒ Q ( u , v , w ) = 2 ( x − y + z ) 2 − 5 y 2 + ( 2 y + z ) 2 = = 2 u 2 − 5 v 2 + w 2 . \begin{array}{l}
\left( \begin{array}{c} u \\ v \\ w \end{array} \right) = \left( \begin{array}{c} x - y + z \\ y \\ 2y + z \end{array} \right) \Rightarrow Q(u, v, w) = 2(x - y + z)^2 - 5y^2 + (2y + z)^2 = \\
= 2u^2 - 5v^2 + w^2.
\end{array} ⎝ ⎛ u v w ⎠ ⎞ = ⎝ ⎛ x − y + z y 2 y + z ⎠ ⎞ ⇒ Q ( u , v , w ) = 2 ( x − y + z ) 2 − 5 y 2 + ( 2 y + z ) 2 = = 2 u 2 − 5 v 2 + w 2 .
iii)
( u v w ) = ( x − y + z y 2 y + z ) ⇒ { u = ( 1 , − 1 , 1 ) v = ( 0 , 1 , 0 ) w = ( 0 , 2 , 1 ) − principal axes . \left( \begin{array}{c} u \\ v \\ w \end{array} \right) = \left( \begin{array}{c} x - y + z \\ y \\ 2y + z \end{array} \right) \Rightarrow \left\{ \begin{array}{l} u = (1, -1, 1) \\ v = (0, 1, 0) \\ w = (0, 2, 1) \end{array} \right. - \text{principal axes}. ⎝ ⎛ u v w ⎠ ⎞ = ⎝ ⎛ x − y + z y 2 y + z ⎠ ⎞ ⇒ ⎩ ⎨ ⎧ u = ( 1 , − 1 , 1 ) v = ( 0 , 1 , 0 ) w = ( 0 , 2 , 1 ) − principal axes .
iv)
rank ( Q ) = rank ( A ) = rank ( 2 0 0 0 − 5 0 0 0 1 ) = 3 ; \operatorname{rank}(Q) = \operatorname{rank}(A) = \operatorname{rank} \left( \begin{array}{ccc} 2 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 1 \end{array} \right) = 3; rank ( Q ) = rank ( A ) = rank ⎝ ⎛ 2 0 0 0 − 5 0 0 0 1 ⎠ ⎞ = 3 ; Q ( u , v , w ) = 2 u 2 − 5 v 2 + w 2 ⇒ sign ( Q ) = 2 − 1 = 1. Q(u, v, w) = 2u^2 - 5v^2 + w^2 \Rightarrow \operatorname{sign}(Q) = 2 - 1 = 1. Q ( u , v , w ) = 2 u 2 − 5 v 2 + w 2 ⇒ sign ( Q ) = 2 − 1 = 1.
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#339914 on Dec 2023