Answer to Question #238277 in Linear Algebra for Anuj

Question #238277

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
2021-11-03T15:11:19-0400

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

"Reduce the quadratic form "8x^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= "\\begin{bmatrix}\n 8&-6& 2 \\\\\n -6&7& -4\\\\\n2&-4&3\n\\end{bmatrix}"


Characteristic equation

det "||A-\\lambda\\Iota||=0"


det "\\begin{Vmatrix}\n 8-\\lambda&-6& 2\\\\\n -6&7-\\lambda & -4\\\\\n2&-4&3-\\lambda\n\\end{Vmatrix}" =0...(i)


Characteristic polynomial

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

Where "D_1=" sum of main diagonal element

"=8+7+3=18"


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


"D_2=\\begin{vmatrix}\n 7& -4\\\\\n -4& 3\n\\end{vmatrix}" + "\\begin{vmatrix}\n 8 & 2 \\\\\n 2 & 3\n\\end{vmatrix}" +


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


"D_3=" Det "|A|" = "8(21-16)-"

-"6(-18+8)+2(56-36)" =


Characteristic equation is

"\\lambda^3-18\\lambda^2+45\\lambda=0"

Solving for "\\lambda;"

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


Finding Eigenvector

Case 1 "\\lambda=0"

"rref \\>\\begin{pmatrix}\n 8&-6 & 2\\\\\n -6&7 & -4\\\\\n2&-4&3\n\\end{pmatrix}" =


"\\begin{pmatrix}\n 1&0& \\frac{1}{2} \\\\\n 0&1 & -1\\\\\n0&0&0\n\\end{pmatrix}"


"\\begin{pmatrix}\n 1&0 & \\frac{-1}{2}\\\\\n 0&1 & -1\\\\\n0&0&0\n\\end{pmatrix}\\begin{pmatrix}\n x \\\\\n y \\\\\nz\n\\end{pmatrix}" =


"\\implies\\>x=\\frac{1}{2}z"

"y=z"


Vector ="\\begin{pmatrix}\n \\frac{1}{2} \\\\\n 1 \\\\\n1\n\\end{pmatrix}" ="\\begin{pmatrix}\n 1 \\\\\n 2\\\\\n2\n\\end{pmatrix}"


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



"\\begin{pmatrix}\n 5&-6 & 2 \\\\\n -6&4 & -4\\\\\n2&-4&0\n\\end{pmatrix}\\begin{pmatrix}\n x \\\\\n y \\\\\nz\n\\end{pmatrix}" =0


Using rref of the matrix


"\\begin{pmatrix}\n 1&0& 1\\\\\n 0&1&\\frac{1}{2} \\\\\n0&0&0\n\\end{pmatrix}" "\\begin{pmatrix}\n x\\\\\n y \\\\\nz\n\\end{pmatrix}" =0


"\\implies\\>x=-z"

"y=\\>\\frac{-1}{2}z" let "z=1"


Vector is "\\begin{pmatrix}\n -1 \\\\\n \\frac{-1}{2} \\\\\n1\n\\end{pmatrix}" ="\\begin{pmatrix}\n -2 \\\\\n -1\\\\\n2\n\\end{pmatrix}"


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


"\\begin{pmatrix}\n -7&-6 & 2 \\\\\n -6&-8& -4\\\\\n2&-4&-12\n\\end{pmatrix}\\begin{pmatrix}\n x \\\\\n y \\\\\nz\n\\end{pmatrix}" =0,using rref form


"\\begin{pmatrix}\n 1&0& -2 \\\\\n 0&1&2 \\\\\n0&0&0\n\\end{pmatrix}\\begin{pmatrix}\n x\\\\\n y \\\\\nz\n\\end{pmatrix}" =0

"\\implies \\>x=2z"

"y= -2z" let "z=1"


Vector is "\\begin{pmatrix}\n 2 \\\\\n -2 \\\\\n1\n\\end{pmatrix}"


Orthonormal matrix from unit eigenvectors



"Q=\\frac{1}{3}" "\\begin{pmatrix}\n 1&-2& 2 \\\\\n 2&-1 & -2\\\\\n2&2&1\n\\end{pmatrix}"


Diagonalizing matrix


"D=Q^TAQ="

"\\frac{1}{9}" "\\begin{pmatrix}\n 1&2 & 2 \\\\\n -2&-1 & 2\\\\\n2&-2&1\n\\end{pmatrix}\\begin{pmatrix}\n 8&-6 & 2 \\\\\n -6&7 & -4\\\\\n2&-4&3\n\\end{pmatrix}\\begin{pmatrix}\n 1&-2 & 2 \\\\\n 2&-1 & -2\\\\\n2&2&1\n\\end{pmatrix}"


="\\begin{pmatrix}\n 0&0 & 0\\\\\n 0&3 & 0\\\\\n0&0&15\n\\end{pmatrix}"


Orthogonal transformation reduce the quadratic form to canonical form


"3y_2^2+15y_3^2"


Nature of quadratic form


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


Therefore it is positive semi-definate




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