Answer to Question #221023 in Linear Algebra for anuj

Question #221023

Reduce the quadratic form 3x2 + 2y2 + 3z2 - 2xy - 2yz into a canonical

form using an orthogonal transformation.


1
Expert's answer
2021-07-28T17:45:15-0400

The matrix of this quadratic form is

"\\begin{pmatrix}\n 3 & -1 & 0\\\\\n -1 & 2 &-1\\\\\n0 & -1 &3\n\\end{pmatrix}"

We have to find its eigenvalues and eigenvectors.

"\\det\\begin{vmatrix}\n 3-x & -1 & 0\\\\\n -1 & 2-x &-1\\\\\n0 & -1 & 3-x\n\\end{vmatrix}=(3-x)^2(2-x)-2(3-x)"

"=(3-x)((3-x)(2-x)-2)=(3-x)(x^2-5x+4)=(3-x)(x-1)(x-4)"

Therefore, the eigenvalues are 1, 3 and 4.

1) Consider the eigenvalue 1:

"\\begin{pmatrix}\n 3-1 & -1 & 0\\\\\n -1 & 2-1 &-1\\\\\n0 & -1 & 3-1\n\\end{pmatrix}\n\\begin{pmatrix}\n x\\\\\n y\\\\\nz\n\\end{pmatrix}=\n\\begin{pmatrix}\n 0\\\\\n 0\\\\\n0\n\\end{pmatrix}"

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

"2x-y=0", "2z-y=0", therefore, the eigenvector is

"\\begin{pmatrix}\n x\\\\\n y\\\\\nz\n\\end{pmatrix}=\n\\begin{pmatrix}\n 2\\\\\n 1\\\\\n2\n\\end{pmatrix}"

Its norm is "\\sqrt{2^2+1^2+2^2}=3". Therefore, the normalized eigenvector, corressponding to the eigenvalue 1, is "\\begin{pmatrix}\n x\\\\\n y\\\\\nz\n\\end{pmatrix}=\n\\begin{pmatrix}\n 2\/3\\\\\n 1\/3\\\\\n2\/3\n\\end{pmatrix}".

2) Consider the eigenvalue 3:

"\\begin{pmatrix}\n 3-3 & -1 & 0\\\\\n -1 & 2-3 &-1\\\\\n0 & -1 & 3-3\n\\end{pmatrix}\n\\begin{pmatrix}\n x\\\\\n y\\\\\nz\n\\end{pmatrix}=\n\\begin{pmatrix}\n 0\\\\\n 0\\\\\n0\n\\end{pmatrix}"

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

"-y=0", "-x-y-z=0", therefore, the eigenvector is

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

Its norm is "\\sqrt{1^2+0^2+(-1)^2}=\\sqrt{2}". Therefore, the normalized eigenvector, corressponding to the eigenvalue 3, is "\\begin{pmatrix}\n 1\/\\sqrt{2}\\\\\n 0\\\\\n-1\/\\sqrt{2}\n\\end{pmatrix}".

3) Consider the eigenvalue 4:

"\\begin{pmatrix}\n 3-4 & -1 & 0\\\\\n -1 & 2-4 &-1\\\\\n0 & -1 & 3-4\n\\end{pmatrix}\n\\begin{pmatrix}\n x\\\\\n y\\\\\nz\n\\end{pmatrix}=\n\\begin{pmatrix}\n 0\\\\\n 0\\\\\n0\n\\end{pmatrix}"

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

"-x-y=0", "-y-z=0", therefore, the eigenvector is

"\\begin{pmatrix}\n x\\\\\n y\\\\\nz\n\\end{pmatrix}=\n\\begin{pmatrix}\n 1\\\\\n -1\\\\\n1\n\\end{pmatrix}"

Its norm is "\\sqrt{1^2+(-1)^2+1^2}=\\sqrt{3}". Therefore, the normalized eigenvector, corressponding to the eigenvalue 3, is "\\begin{pmatrix}\n 1\/\\sqrt{3}\\\\\n -1\/\\sqrt{3}\\\\\n1\/\\sqrt{3}\n\\end{pmatrix}".

By the construction we have

"\\begin{pmatrix}\n 3 & -1 & 0\\\\\n -1 & 2 &-1\\\\\n0 & -1 &3\n\\end{pmatrix}\\begin{pmatrix}\n 2\/ 3 & 1\/\\sqrt{2} & 1\/\\sqrt{3}\\\\\n 1\/3 & 0 &-1\/\\sqrt{3}\\\\\n2\/3 & -1\/\\sqrt{2} &1\/\\sqrt{3}\n\\end{pmatrix}="

"=\\begin{pmatrix}\n 2\/ 3 & 1\/\\sqrt{2} & 1\/\\sqrt{3}\\\\\n 1\/3 & 0 &-1\/\\sqrt{3}\\\\\n2\/3 & -1\/\\sqrt{2} &1\/\\sqrt{3}\n\\end{pmatrix}\\begin{pmatrix}\n 1 & 0 & 0\\\\\n 0 & 3 &0\\\\\n0 & 0 &4\n\\end{pmatrix}"

and

"3x^2+2y^2+3z^2-2xy-2yz=\\begin{pmatrix}\n x &y & z\n\\end{pmatrix}\\begin{pmatrix}\n 3 & -1 & 0\\\\\n -1 & 2 &-1\\\\\n0 & -1 &3\n\\end{pmatrix}\\begin{pmatrix}\n x\\\\\n y\\\\\nz\n\\end{pmatrix}="

"\\begin{pmatrix}\n x &y & z\n\\end{pmatrix}\\begin{pmatrix}\n 2\/ 3 & 1\/\\sqrt{2} & 1\/\\sqrt{3}\\\\\n 1\/3 & 0 &-1\/\\sqrt{3}\\\\\n2\/3 & -1\/\\sqrt{2} &1\/\\sqrt{3}\n\\end{pmatrix}\\begin{pmatrix}\n 1 & 0 & 0\\\\\n 0 & 3 &0\\\\\n0 & 0 &4\n\\end{pmatrix}\\begin{pmatrix}\n 2\/ 3 & 1\/\\sqrt{2} & 1\/\\sqrt{3}\\\\\n 1\/3 & 0 &-1\/\\sqrt{3}\\\\\n2\/3 & -1\/\\sqrt{2} &1\/\\sqrt{3}\n\\end{pmatrix}^{-1}\\begin{pmatrix}\n x\\\\\n y\\\\\nz\n\\end{pmatrix}"

But the matrix

"\\begin{pmatrix}\n 2\/ 3 & 1\/\\sqrt{2} & 1\/\\sqrt{3}\\\\\n 1\/3 & 0 &-1\/\\sqrt{3}\\\\\n2\/3 & -1\/\\sqrt{2} &1\/\\sqrt{3}\n\\end{pmatrix}" is orthogonal, since it is composed with the orthonormal eigenvectors of the symmetric matrice. therefore

"\\begin{pmatrix}\n 2\/ 3 & 1\/\\sqrt{2} & 1\/\\sqrt{3}\\\\\n 1\/3 & 0 &-1\/\\sqrt{3}\\\\\n2\/3 & -1\/\\sqrt{2} &1\/\\sqrt{3}\n\\end{pmatrix}^{-1}=\\begin{pmatrix}\n 2\/ 3 & 1\/\\sqrt{2} & 1\/\\sqrt{3}\\\\\n 1\/3 & 0 &-1\/\\sqrt{3}\\\\\n2\/3 & -1\/\\sqrt{2} &1\/\\sqrt{3}\n\\end{pmatrix}^{T}"

"=\\begin{pmatrix}\n 2\/ 3 & 1\/\\sqrt{2} & 1\/\\sqrt{3}\\\\\n 1\/3 & 0 &-1\/\\sqrt{3}\\\\\n2\/3 & -1\/\\sqrt{2} &1\/\\sqrt{3}\n\\end{pmatrix}"

Put

"\\begin{pmatrix}\n 2\/ 3 & 1\/\\sqrt{2} & 1\/\\sqrt{3}\\\\\n 1\/3 & 0 &-1\/\\sqrt{3}\\\\\n2\/3 & -1\/\\sqrt{2} &1\/\\sqrt{3}\n\\end{pmatrix}\\begin{pmatrix}\n x\\\\\n y\\\\\nz\n\\end{pmatrix} = \\begin{pmatrix}\n X\\\\\n Y\\\\\nZ\n\\end{pmatrix}"

then

"3x^2+2y^2+3z^2-2xy-2yz=(X\\,Y\\,Z)\\begin{pmatrix}\n 1 & 0 & 0\\\\\n 0 & 3 &0\\\\\n0 & 0 &4\n\\end{pmatrix}\\begin{pmatrix}\n X\\\\\n Y\\\\\nZ\n\\end{pmatrix}=X^2+3Y^2+4Z^2"

And we reduce the given quadratic form to the canonical form by orthogonal transformation.


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