DEFINATION-OF-QUADRATIC-FORM-A-Quadratic-form-is-a-homogeneous-polynomial-of-degree-two-in-multiple-variable-Q-X-T-AX-Here-Q-Quadratic-form-ax-2-by-2-cz

Question Number 212028 by siva12345 last updated on 27/Sep/24

D E F I N A T I O N O F Q U A D R A T I C F O R M :

A Q u a d r a t i c f o r m i s a h o m o g e n e o u s p o l y n o m i a l o f d e g r e e t w o i n m u l t i p l e v a r i a b l e .

Q = X^{T} A X

H e r e Q = Q u a d r a t i c f o r m .

{a x}^{2} + {b y}^{2} + {c z}^{2} + 2 h x y + 2 f y z + 2 g z x = 0

B y u s i n g t h e s e Q = X^{T} A X [w e c a n w r i t e m a t r i x A]

A = [\begin{matrix} x \\ y \\ z \end{matrix}] [\begin{matrix} a h g \\ h b f \\ g f c \end{matrix}] [\begin{matrix} x & y & z \end{matrix}]

B y u s i n g t h e c h a r a c t e r i s t i c e q u a t i o n o f m a t r i x ∣ A - λ I ∣ = 0

T o g e t e i g e n e q u a t i o n t o c a n s o l v e e i g e n e q u a t i o n s t o g e t e i g e n v a l u e s .

S u p p o s e s u b s t i t u t e t h e e i g w n v a l u e i n ∣ A - 1 λ I ∣ X = 0

A f t e r r o w t r a n s f o r m a t i o n w e c a n g e t x_{1}, x_{2}, x_{3}

P = [e_{1} e_{2} e_{3}] ∴ {H e r e p i s a m o d a l m a t r i x}

S i n c e p i s a o r t h o g o n a l m a t r i x p^{- 1} = p

D = P^{T} A P W h e r e D i s d i a g o n a l m a t r i x .

T h e q u a d r a t i c f o r m c a n b e r e d u c e i n t o n o r m a l f o r m X = Y^{T} D Y .

H e r e :

I n d e x = T h e n u m b e r o f p o s i t i v e t e r m s i n i t s c a n o n i c a l f o r m

S i g n a t u r e = T h e d i f f e r e n c e o f p o s i t i v e a n d n e g a t i v e t e r m i n t h e c a n o n i c a l f o r m

I f a l l λ > 0 \Rightarrow p o s i t i v e d e f i n i t e .

I f a l l λ < 0 \Rightarrow n e g a t i v e d e f i n i t e .

I f a l l λ \geq 0 a t l e a s t o n e λ = 0 \Rightarrow p o s i t i v e s e m i d e f i n i t e .

I f a l l λ \leq 0 a t l e s t o n e λ = 0 \Rightarrow n e g a t i v e s e m i d e f i n i t e .

I f s o m e λ a r e p o s i t i v e a n d s o m e λ a r e n e g a t i v e \Rightarrow i n d e f i n i t e .

X = P Y i s u s e d t r a n s f o r m a t i o n o f q u a d r a t i c f o r m t o n o r m a l f o r m (o r) c a n o n i c a l f o r m .

E g :

(Q) R e d u c e t h e q u a d r a t i c f o r m 3 x^{2} + 2 y^{2} + 3 z^{2} - 2 x y - 2 y z = 0 t o c a n o n i c a l f o r m b y o r t h o g o n a l t r a n s f o r m a t i o n

a n d a l s o f i n d r a n k, i n d e x, n a t u r e a n d s i g n a t u r e .

S o l : T h e g i v e n e q u a t i o n i s 3 x^{2} + 2 y^{2} + 3 z^{2} - 2 x y - 2 y z = 0

3 x x + 2 y y + 3 z z - x y - x y - y z - y z = 0

w e k n o w t h a t Q = X^{T} A X

= [\begin{matrix} X \\ Y \\ Z \end{matrix}] [\begin{matrix} 3 - 1 0 \\ - 1 2 - 1 \\ 0 - 1 3 \end{matrix}] [\begin{matrix} X & Y & Z \end{matrix}]

T h e c o r r e s p o n d i n g s y m m e t r i c m a t r i x A = [\begin{matrix} 3 - 1 0 \\ - 1 2 - 1 \\ 0 - 1 3 \end{matrix}]

T o f i n d t h e e i g e n v a l u e s w e c a n u s e t h e c h a r e t e r i s t i c e a u a t i o n o f m a t r i x A i s ∣ A - λ I ∣= 0

[\begin{matrix} 3 - λ - 1 0 \\ - 1 2 - λ - 1 \\ 0 - 1 3 - λ \end{matrix}] = 0

(3 - λ) [(2 - λ) (3 - λ) - 1)] + 1 [- 1 (3 - λ)] = 0

(3 - λ) (- 5 λ + λ^{2} + 5) + (λ - 3) = 0

λ^{3} - 8 λ^{2} + 19 λ - 12 = 0

λ = 1, 3, 4

C a s e i :

λ = 1 s u b s t i t u t e i n [A - λ I] X = 0

[\begin{matrix} 2 - 1 0 \\ - 1 1 - 1 \\ 0 - 1 2 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]

R_{2} \Rightarrow R_{2} + 2 R_{1}

[\begin{matrix} 2 - 1 0 \\ 0 1 - 2 \\ 0 - 1 2 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]

R_{3} \Rightarrow R_{3} + R_{2}

[\begin{matrix} 2 - 1 0 \\ 0 1 - 2 \\ 0 0 0 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]

2 x_{1} - x_{2} = 0

x_{2} - 2 x_{3} = 0

x_{3} = k

x_{2} = 2 k

x_{1} = k

x_{1} = [\begin{matrix} 1 \\ 2 \\ 1 \end{matrix}] k

C a s e i i :

λ = 3 T o s u b s t i t i t e λ v a l u e i n [A - λ I] X = 0

[\begin{matrix} 0 - 1 0 \\ - 1 - 1 - 1 \\ 0 - 1 0 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]

(R_{1} \Leftrightarrow R_{2})

[\begin{matrix} - 1 - 1 - 1 \\ 0 - 1 0 \\ 0 - 1 0 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]

R_{3} \Rightarrow R_{3} - R_{2}

[\begin{matrix} - 1 - 1 - 1 \\ 0 - 1 0 \\ 0 0 0 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x 3 \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]

x_{2} = 0

- x_{1} - x_{2} - x_{3} = 0

x_{3} = k

x_{1} = - k

x_{2} = [\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}] k

C a s e i i i :

λ = 4 T o s u b s t i t u t e i n e q u a t i o n

[A - λ I] X = 0

[\begin{matrix} - 1 - 1 0 \\ - 1 - 2 - 1 \\ 0 - 1 - 1 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]

R_{2} \Rightarrow R_{2} - R_{1}

[\begin{matrix} - 1 - 1 0 \\ 0 - 1 - 1 \\ 0 - 1 - 1 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]

R_{3} \Rightarrow R_{3} - R_{2}

[\begin{matrix} - 1 - 1 0 \\ 0 - 1 - 1 \\ 0 0 0 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]

- x_{1} - x_{2} = 0

- x_{3} - x_{3} = 0

x_{3} = k

x_{2} = - k

x_{1} = k

x_{3} = [\begin{matrix} 1 \\ - 1 \\ 1 \end{matrix}] k

w e c a n o b s e r v e t h a t v e c t o r s a r e o r t h o g o n a l l y m u t u a l l y w e n o r m a l i z e t h i s v e c t o r s a n d o b t a i n .

e_{1} = [\begin{matrix} \frac{1}{\sqrt{6}} \\ \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \end{matrix}] e_{2} = [\begin{matrix} \frac{- 1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \end{matrix}]

e_{3} = [\begin{matrix} \frac{1}{\sqrt{3}} \\ \frac{- 1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \end{matrix}]

l e t P = t h e m o d a l m a t r i x i n n o r m a l i z e d f o r m

P = [e_{1} e_{2} e_{3}] = [\begin{matrix} \frac{1}{\sqrt{6}} \frac{- 1}{\sqrt{2}} \frac{1}{\sqrt{3}} \\ \frac{2}{\sqrt{6}} 0 \frac{- 1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} \frac{1}{\sqrt{2}} \frac{- 1}{\sqrt{3}} \end{matrix}]

S i n c e p i s a o r t h o g o n a l m a t r i x P^{- 1} = P^{T}

S o D = P^{T} A P w h e r e D i s d i a g n o l m a t r i x

= [\begin{matrix} \frac{1}{\sqrt{6}} \frac{- 2}{\sqrt{6}} \frac{1}{\sqrt{6}} \\ \frac{- 1}{\sqrt{2}} 0 \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} \frac{- 1}{\sqrt{3}} \frac{- 1}{\sqrt{3}} \end{matrix}] [\begin{matrix} 3 - 1 0 \\ - 1 2 - 1 \\ 0 - 1 3 \end{matrix}] [\begin{matrix} \frac{1}{\sqrt{6}} - \frac{1}{\sqrt{2}} \frac{1}{\sqrt{3}} \\ \frac{2}{\sqrt{6}} 0 \frac{- 1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}} \end{matrix}]

D = [\begin{matrix} 1 0 0 \\ 0 3 0 \\ 0 0 4 \end{matrix}]

T h e q u a d r a t i c f o r m c a n b e r e d u c e t o n o r m a l f o r m Y^{T} D Y

Y^{T} D Y = [y_{1} y_{2} y_{3}] [\begin{matrix} 1 0 0 \\ 0 3 0 \\ 0 0 4 \end{matrix}] [\begin{matrix} y_{1} \\ y_{2} \\ y_{3} \end{matrix}]

y_{1}^{2} + 3 y_{2}^{2} + 4 y_{3}^{2} = 0

R a n k = 3 I n d e x = 3

S i g n a t u r e = 3 N a t u r e = P o s i t i v e d e f i n i t e

B y o r t h o g o n a l t f a n s f o r m a t i o n X = P Y

[\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} \frac{1}{\sqrt{6}} \frac{- 1}{\sqrt{2}} \frac{1}{\sqrt{3}} \\ \frac{2}{\sqrt{6}} 0 \frac{- 1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} \frac{1}{\sqrt{2}} \frac{1}{\sqrt{3}} \end{matrix}] [\begin{matrix} y_{1} \\ y_{2} \\ y_{3} \end{matrix}]

x_{1} = \frac{y_{1}}{\sqrt{6}} - \frac{y_{2}}{\sqrt{2}} + \frac{y_{3}}{\sqrt{3}} .

x_{2} = \frac{2 y_{1}}{\sqrt{6}} - \frac{y_{3}}{\sqrt{3 .}}

x_{3} = \frac{y_{1}}{\sqrt{6}} + \frac{y_{2}}{\sqrt{2}} + \frac{y_{3}}{\sqrt{3 .}}

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