| 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^2 +2hxy+2fyz+2gzx=0
By using these Q=X^T AX [we can write matrix A]
A= [(x),(y),(z) ] [((a h g)),((h b f)),((g f c)) ] [(x,y,z) ]
By using the characteristic equation of matrix ∣A−λI∣ =0
To get eigen equation to can solve eigen equations to get eigen values.
Suppose substitute the eigwn value in ∣A−1λI∣X =0
After row transformation we can get x_1 ,x_2 ,x_3
P =[e_(1 ) e_2 e_3 ] ∴ {Here p is a modal matrix}
Since p is a orthogonal matrix p^(−1) =p
D=P^T AP Where D is diagonal matrix.
The quadratic form can be reduce into normal form X = Y^T DY.
Here:
Index = The number of positive terms in its canonical form
Signature = The difference of positive and negative term in the canonical form
If all λ>0⇒ positive definite.
If all λ<0⇒ negative definite.
If all λ≥0 atleast one λ= 0⇒positive semidefinite.
If all λ≤ 0 atlest one λ= 0⇒negative semidefinite.
If some λ are positive and some λ are negative⇒indefinite.
X = PY is used transformation of quadratic form to normal form (or) canonical form.
Eg:
(Q) Reduce the quadratic form 3x^2 +2y^2 +3z^2 −2xy−2yz = 0 to canonical form by orthogonal transformation
and also find rank, index, nature and signature.
Sol: The given equation is 3x^2 +2y^2 +3z^2 −2xy−2yz=0
3xx+2yy+3zz−xy−xy−yz−yz=0
we know that Q=X^T AX
= [(X),(Y),(Z) ] [(( 3 −1 0)),((−1 2 −1)),(( 0 −1 3)) ] [(X,Y,Z) ]
The corresponding symmetric matrix A= [(( 3 −1 0)),((−1 2 −1)),(( 0 −1 3)) ]
To find the eigen values we can use the chareteristic eauation of matrix A is ∣A−λI∣= 0
[(( 3−λ −1 0)),((−1 2−λ −1)),(( 0 −1 3−λ)) ]= 0
(3−λ)[(2−λ)(3−λ)−1)]+1[−1(3−λ)]= 0
(3−λ)(−5λ+λ^2 +5)+(λ−3)= 0
λ^3 −8λ^2 +19λ−12= 0
λ=1,3,4
Case i:
λ=1 substitute in [A−λI]X = 0
[(( 2 −1 0)),((−1 1 −1)),(( 0 −1 2)) ] [(x_1 ),(x_2 ),(x_3 ) ]= [(0),(0),(0) ]
R_2 ⇒R_2 +2R_1
[(( 2 −1 0 )),(( 0 1 −2)),(( 0 −1 2)) ] [(x_1 ),(x_2 ),(x_3 ) ] = [(0),(0),(0) ]
R_3 ⇒R_3 +R_2
[((2 −1 0 )),((0 1 −2)),((0 0 0)) ] [(x_1 ),(x_2 ),(x_3 ) ] = [(0),(0),(0) ]
2 x_(1 ) −x_2 =0
x_2 −2x_3 =0
x_3 = k
x_2 = 2k
x_1 = k
x_1 = [(1),(2),(1) ]k
Case ii:
λ = 3 To substitite λ value in [A−λI]X = 0
[(( 0 −1 0)),((−1 −1 −1)),(( 0 −1 0)) ] [(x_1 ),(x_2 ),(x_3 ) ]= [(0),(0),(0) ]
(R_1 ⇔R_2 )
[(( −1 −1 −1)),(( 0 −1 0)),(( 0 −1 0)) ] [(x_1 ),(x_2 ),(x_3 ) ]= [((0 )),(0),(0) ]
R_3 ⇒R_3 −R_2
[(( −1 −1 −1)),(( 0 −1 0)),(( 0 0 0)) ] [(x_1 ),(x_2 ),((x3)) ] = [(0),(0),(0) ]
x_2 = 0
−x_1 −x_2 −x_(3 ) =0
x_3 = k
x_(1 ) = −k
x_2 = [((−1)),(( 0)),(( 1)) ]k
Case iii:
λ = 4 To substitute in equation
[ A−λI ]X = 0
[(( −1 −1 0)),(( −1 −2 −1)),(( 0 −1 −1)) ] [(x_1 ),(x_2 ),(x_3 ) ] = [(0),(0),(0) ]
R_2 ⇒R_2 −R_1
[(( −1 −1 0)),(( 0 −1 −1)),(( 0 −1 −1)) ] [(x_1 ),(x_2 ),(x_3 ) ] = [(0),(0),(0) ]
R_3 ⇒R_3 −R_2
[(( −1 −1 0)),(( 0 −1 −1)),(( 0 0 0)) ] [(x_1 ),(x_2 ),(x_3 ) ] = [(0),(0),(0) ]
−x_1 −x_(2 ) = 0
−x_3 −x_3 = 0
x_(3 ) = k
x_2 = −k
x_1 = k
x_(3 ) = [(( 1)),((−1)),(( 1 )) ]k
we can observe that vectors are orthogonally mutually we normalize this vectors and obtain.
e_1 = [((1/( (√6)))),((2/( (√6)))),((1/( (√6)))) ] e_2 = [(((−1)/( (√2)))),(( 0)),(( (1/( (√2))))) ]
e_3 = [((1/( (√3)))),(((−1)/( (√3)))),((1/( (√3)))) ]
let P = the modal matrix in normalized form
P = [e_(1 ) e_2 e_3 ] = [(((1/( (√6))) ((−1)/( (√(2 )))) (1/( (√3))))),(((2/( (√(6 )) )) 0 ((−1)/( (√3))))),(((1/( (√6))) (1/( (√2))) ((−1)/( (√3))))) ]
Since p is a orthogonal matrix P^(−1 ) = P^T
So D = P^T AP where D is diagnol matrix
= [(((1/( (√6))) ((−2)/( (√6))) (1/( (√6))))),((((−1)/( (√2))) 0 (1/( (√2))))),(((1/( (√3))) ((−1)/( (√3))) ((−1)/( (√3))))) ] [(( 3 −1 0)),((−1 2 −1)),(( 0 −1 3)) ] [(((1/( (√(6 )))) −(1/( (√2))) (1/( (√3))))),(((2/( (√6))) 0 ((−1)/( (√3))))),(((1/( (√6))) (1/( (√2))) −(1/( (√3))))) ]
D = [((1 0 0 )),((0 3 0)),((0 0 4)) ]
The quadratic form can be reduce to normal form Y^T DY
Y^( T) DY = [y_1 y_(2 ) y_3 ] [((1 0 0)),((0 3 0)),((0 0 4)) ] [(y_1 ),(y_2 ),(y_3 ) ]
y_1 ^2 +3y_2 ^2 +4y_3 ^(2 ) = 0
Rank = 3 Index = 3
Signature= 3 Nature=Positive definite
By orthogonal tfansformation X = PY
[(x_1 ),(x_2 ),(x_3 ) ] = [(((1/( (√6))) ((−1)/( (√2))) (1/( (√3))))),(((2/( (√6))) 0 ((−1)/( (√3))))),(((1/( (√(6 )))) (1/( (√2))) (1/( (√3))))) ] [(y_1 ),(y_2 ),(y_3 ) ]
x_(1 ) = (y_1 /( (√6)))−(y_2 /( (√2)))+(y_3 /( (√3))).
x_2 = ((2y_1 )/( (√6)))−(y_3 /( (√(3.))))
x_3 = (y_1 /( (√6)))+(y_2 /( (√2)))+(y_3 /( (√(3.))))
o
|
