Question #246097
Reduce the quadratic form


2 2 2 8 7 3 12 – 8 4 x y z xy yz zx    


to the canonical form


through an orthogonal transformation and hence show that it


is positive Semi-definite.
1
Expert's answer
2021-11-07T16:55:15-0500

Assuming typo error, and modifying the condition of the question into

"Reduce the quadratic form 8x2+7y2+3z212xy8yz+4zx8x^2+7y^2+3z^2-12xy-8yz+4zx

to the canonical form through orthogonal transformation and hence show it is positive semi-definate"


Solution

Converting to matrix form


A= [862674243]\begin{bmatrix} 8&-6& 2 \\ -6&7& -4\\ 2&-4&3 \end{bmatrix}



Characteristic equation

det AλI=0||A-\lambda\Iota||=0


det 8λ6267λ4243λ\begin{Vmatrix} 8-\lambda&-6& 2\\ -6&7-\lambda & -4\\ 2&-4&3-\lambda \end{Vmatrix} =0...(i)


Characteristic polynomial

λ3D1λ2+D2λD3=0\lambda^3-D_1\lambda^2+D_2\lambda-D_3=0

Where D1=D_1= sum of main diagonal element

=8+7+3=18=8+7+3=18

D2=D_2= sum of minors of the main diagonal element


D2=7443D_2=\begin{vmatrix} 7& -4\\ -4& 3 \end{vmatrix} + 8223\begin{vmatrix} 8 & 2 \\ 2 & 3 \end{vmatrix} +


8667\begin{vmatrix} 8 & -6\\ -6& 7 \end{vmatrix} =5+20+20=45=5+20+20=45


D3=D_3= Det A|A| = 8(2116)8(21-16)-

-6(18+8)+2(5636)6(-18+8)+2(56-36) =


Characteristic equation is

λ318λ2+45λ=0\lambda^3-18\lambda^2+45\lambda=0

Solving for λ;\lambda;


λ=0,λ=3,λ=15\lambda=0, \>\lambda=3,\>\lambda=15


Finding Eigenvector

Case 1 λ=0\lambda=0

rref(862674243)rref \>\begin{pmatrix} 8&-6 & 2\\ -6&7 & -4\\ 2&-4&3 \end{pmatrix} =


(1012011000)\begin{pmatrix} 1&0& \frac{1}{2} \\ 0&1 & -1\\ 0&0&0 \end{pmatrix}


(1012011000)(xyz)\begin{pmatrix} 1&0 & \frac{-1}{2}\\ 0&1 & -1\\ 0&0&0 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} =


    x=12z\implies\>x=\frac{1}{2}z

y=zy=z


Vector =(1211)\begin{pmatrix} \frac{1}{2} \\ 1 \\ 1 \end{pmatrix} =(122)\begin{pmatrix} 1 \\ 2\\ 2 \end{pmatrix}


Case 2 λ=3,\lambda =3, substituting λ=3\lambda=3 in.....(i)



(562644240)(xyz)\begin{pmatrix} 5&-6 & 2 \\ -6&4 & -4\\ 2&-4&0 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} =0


Using ref of the matrix


(1010112000)\begin{pmatrix} 1&0& 1\\ 0&1&\frac{1}{2} \\ 0&0&0 \end{pmatrix} (xyz)\begin{pmatrix} x\\ y \\ z \end{pmatrix} =0


    x=z\implies\>x=-z

y=12zy=\>\frac{-1}{2}z let z=1z=1


Vector is (1121)\begin{pmatrix} -1 \\ \frac{-1}{2} \\ 1 \end{pmatrix} =(212)\begin{pmatrix} -2 \\ -1\\ 2 \end{pmatrix}


Case 3, λ=15,\lambda=15, substituting λ=15\lambda=15 in .....(i)


(7626842412)(xyz)\begin{pmatrix} -7&-6 & 2 \\ -6&-8& -4\\ 2&-4&-12 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} =0,using rref form


(102012000)(xyz)\begin{pmatrix} 1&0& -2 \\ 0&1&2 \\ 0&0&0 \end{pmatrix}\begin{pmatrix} x\\ y \\ z \end{pmatrix} =0

    x=2z\implies \>x=2z

y=2zy= -2z let z=1z=1

Vector is (221)\begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}


Orthonormal matrix from unit eigenvectors



Q=13Q=\frac{1}{3} (122212221)\begin{pmatrix} 1&-2& 2 \\ 2&-1 & -2\\ 2&2&1 \end{pmatrix}

Diagonalizing matrix


D=QTAQ=D=Q^TAQ=

19\frac{1}{9} (122212221)(862674243)(122212221)\begin{pmatrix} 1&2 & 2 \\ -2&-1 & 2\\ 2&-2&1 \end{pmatrix}\begin{pmatrix} 8&-6 & 2 \\ -6&7 & -4\\ 2&-4&3 \end{pmatrix}\begin{pmatrix} 1&-2 & 2 \\ 2&-1 & -2\\ 2&2&1 \end{pmatrix}


=(0000300015)\begin{pmatrix} 0&0 & 0\\ 0&3 & 0\\ 0&0&15 \end{pmatrix}


Orthogonal transformation reduce the quadratic form to canonical form


3y22+15y323y_2^2+15y_3^2


Nature of quadratic form


Contain a zero, eigenvalue and 2 eigenvalue >1>1


Therefore it is positive semi-definate

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS