Answer to Question #197041 in Linear Algebra for Abhishek

Question

Reduce the quadratic form to a canonical form and find its nature

1) 2xy+2yz+2zx

Answer 1

Given question is 2xy+2yz+2zx

The general equation $a_1x^2+a_2y^2+a_3z^2+a_4 xy+a_5yz+a_6 zx$

For the given equation, $a_1=a_2=a_3=0$

$a_4=2, a_5=2, a_6=2$

So, matrix $A=\begin{bmatrix} 0 & 1 & 1\\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{bmatrix}$

After the matrix operation,

Let $\lambda$ be the eigen value of the equation, so $A-\lambda I=0$

Given question is 2xy+2yz+2zx

The general equation $a_1x^2+a_2y^2+a_3z^2+a_4 xy+a_5yz+a_6 zx$

For the given equation, $a_1=a_2=a_3=0$

$a_4=2, a_5=2, a_6=2$

So, matrix $A=\begin{bmatrix} 0 & 1 & 1\\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{bmatrix}$

Let the eigen value for the equation be $\lambda,$ From the characteristic equation,

$|A-\lambda I|=0$

$\begin{bmatrix} -\lambda & 1 & 1\\ 1 & -\lambda & 1 \\ 1 & 1 & -\lambda \\ \end{bmatrix}=0$

After the expansion of the matrix,

$\Rightarrow -\lambda^3+3\lambda -2=0$

$\Rightarrow \lambda^3-3\lambda +2=0$

$(\lambda +1)(\lambda +1)(\lambda -2)=0$

$\lambda =-1, \lambda =-1, \lambda =2$

Now, substituting the eigen values,

$\lambda = 2,$ $x_1=\begin{bmatrix} 1\\ 1 \\ 1 \\ \end{bmatrix}$

$\lambda=-1, x_2=\begin{bmatrix} -1\\ 1 \\ 0 \\ \end{bmatrix}$

$\lambda=-1, x_3=\begin{bmatrix} 0\\ 1 \\ -1 \\ \end{bmatrix}$

Now, $P=(y_1,y_2,y_2)$

Applying the transpose of matrix P,

$P^t=\begin{bmatrix} y_1\\ y_2 \\ y_3 \end{bmatrix}$

Now, multiplying the P , A and transpose matrix of P

$PAP^t=\begin{bmatrix} y_1 & y_2 & y_3 \\ \end{bmatrix}\begin{bmatrix} -2 & 1 & 1\\ 1 & -1 & 1 \\ 1 & 1 & -1 \\ \end{bmatrix}\begin{bmatrix} y_1\\ y_2 \\ y_3 \end{bmatrix}$

$=\begin{bmatrix} -2y_1 +y_2 + y_3 &y_1-y_2+y_3 & y_1+y_2-y_3 \end{bmatrix}\begin{bmatrix} y_1\\ y_2 \\ y_3 \end{bmatrix}$

$=-2y_1^2+y_1y_2+y_1y_3+y_1y_2-y_2^2+y_2y_3+y_1y_3+y_2y_3-y_3^2$

$=-(y_1^2+y_2^2+y_3^2-2y_1y_2-2y_2y_3-2y_3y_1)$

$=-(y_1-y_2-y_3)^2$

The above equation is in the canonical form.

Rank =3

Index = 1