Answer to Question #220896 in Linear Algebra for Nicky

Question

Reduce the quadratic form 3x2 + 2y2 + 3z2 - 2xy - 2yz into a canonical
form using an orthogonal transformation.

Answer 1

Answer:-

The matrix of this quadratic form is

$\begin{pmatrix} 3 & -1 & 0\\ -1 & 2 &-1\\ 0 & -1 &3 \end{pmatrix}$

We have to find its eigenvalues and eigenvectors.

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

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

Therefore, the eigenvalues are 1, 3 and 4.

1) Consider the eigenvalue 1:

$\begin{pmatrix} 3-1 & -1 & 0\\ -1 & 2-1 &-1\\ 0 & -1 & 3-1 \end{pmatrix} \begin{pmatrix} x\\ y\\ z \end{pmatrix}= \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}$

$\begin{pmatrix} 2 & -1 & 0\\ -1 &1 &-1\\ 0 & -1 & 2 \end{pmatrix} \begin{pmatrix} x\\ y\\ z \end{pmatrix}= \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}$

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

$\begin{pmatrix} x\\ y\\ z \end{pmatrix}= \begin{pmatrix} 2\\ 1\\ 2 \end{pmatrix}$

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

2) Consider the eigenvalue 3:

$\begin{pmatrix} 3-3 & -1 & 0\\ -1 & 2-3 &-1\\ 0 & -1 & 3-3 \end{pmatrix} \begin{pmatrix} x\\ y\\ z \end{pmatrix}= \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}$

$\begin{pmatrix} 0 & -1 & 0\\ -1 &-1 &-1\\ 0 & -1 & 0 \end{pmatrix} \begin{pmatrix} x\\ y\\ z \end{pmatrix}= \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}$

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

$\begin{pmatrix} x\\ y\\ z \end{pmatrix}= \begin{pmatrix} 1\\ 0\\ -1 \end{pmatrix}$

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

3) Consider the eigenvalue 4:

$\begin{pmatrix} 3-4 & -1 & 0\\ -1 & 2-4 &-1\\ 0 & -1 & 3-4 \end{pmatrix} \begin{pmatrix} x\\ y\\ z \end{pmatrix}= \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}$

$\begin{pmatrix} -1 & -1 & 0\\ -1 &-2 &-1\\ 0 & -1 & -1 \end{pmatrix} \begin{pmatrix} x\\ y\\ z \end{pmatrix}= \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}$

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

$\begin{pmatrix} x\\ y\\ z \end{pmatrix}= \begin{pmatrix} 1\\ -1\\ 1 \end{pmatrix}$

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

By the construction we have

$\begin{pmatrix} 3 & -1 & 0\\ -1 & 2 &-1\\ 0 & -1 &3 \end{pmatrix}\begin{pmatrix} 2/ 3 & 1/\sqrt{2} & 1/\sqrt{3}\\ 1/3 & 0 &-1/\sqrt{3}\\ 2/3 & -1/\sqrt{2} &1/\sqrt{3} \end{pmatrix}=$

$=\begin{pmatrix} 2/ 3 & 1/\sqrt{2} & 1/\sqrt{3}\\ 1/3 & 0 &-1/\sqrt{3}\\ 2/3 & -1/\sqrt{2} &1/\sqrt{3} \end{pmatrix}\begin{pmatrix} 1 & 0 & 0\\ 0 & 3 &0\\ 0 & 0 &4 \end{pmatrix}$

and

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

$\begin{pmatrix} x &y & z \end{pmatrix}\begin{pmatrix} 2/ 3 & 1/\sqrt{2} & 1/\sqrt{3}\\ 1/3 & 0 &-1/\sqrt{3}\\ 2/3 & -1/\sqrt{2} &1/\sqrt{3} \end{pmatrix}\begin{pmatrix} 1 & 0 & 0\\ 0 & 3 &0\\ 0 & 0 &4 \end{pmatrix}\begin{pmatrix} 2/ 3 & 1/\sqrt{2} & 1/\sqrt{3}\\ 1/3 & 0 &-1/\sqrt{3}\\ 2/3 & -1/\sqrt{2} &1/\sqrt{3} \end{pmatrix}^{-1}\begin{pmatrix} x\\ y\\ z \end{pmatrix}$

But the matrix

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

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

$=\begin{pmatrix} 2/ 3 & 1/\sqrt{2} & 1/\sqrt{3}\\ 1/3 & 0 &-1/\sqrt{3}\\ 2/3 & -1/\sqrt{2} &1/\sqrt{3} \end{pmatrix}$

Put

$\begin{pmatrix} 2/ 3 & 1/\sqrt{2} & 1/\sqrt{3}\\ 1/3 & 0 &-1/\sqrt{3}\\ 2/3 & -1/\sqrt{2} &1/\sqrt{3} \end{pmatrix}\begin{pmatrix} x\\ y\\ z \end{pmatrix} = \begin{pmatrix} X\\ Y\\ Z \end{pmatrix}$

then

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

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