Answer to Question #153072 in Linear Algebra for Kishore Kris

Question

3. Reduce the quadratic form x 1 ^ 2 +5 x 2 ^ 2 + x 3 ^ 2 +2x 1 x 2 +2x 2 x 3 +6x 3 x 1 to canonical form through an orthogonal transformation and also find its rank, index, signature and nature of the quadratic form.

Answer 1

The matrix of the quadratic form is

$$A=\begin{bmatrix} 1 & 1 & 3 \\ 1 & 5 & 1 \\ 3 & 1 & 1 \\ \end{bmatrix}$$

Find the eigenvalues and eigenvectors of A

$$A-\lambda I=\begin{bmatrix} 1-\lambda & 1 & 3 \\ 1 & 5-\lambda & 1 \\ 3 & 1 & 1-\lambda \\ \end{bmatrix}$$

$$\begin{vmatrix} 1-\lambda & 1 & 3 \\ 1 & 5-\lambda & 1 \\ 3 & 1 & 1-\lambda \\ \end{vmatrix}=(1-\lambda)\begin{vmatrix} 5-\lambda & 1 \\ 1 & 1-\lambda \end{vmatrix}$$

$$-(1)\begin{vmatrix} 1 & 1 \\ 3 & 1-\lambda \end{vmatrix}+(3)\begin{vmatrix} 1 & 5-\lambda \\ 3 & 1 \end{vmatrix}$$

$$=(1-\lambda)^2(5-\lambda)-1+\lambda-1+\lambda+3+3-45+9\lambda$$

$$=-\lambda^3+7\lambda^2-36$$

This is a characteristic polynomial.

Solve the equation

$$-\lambda^3+7\lambda^2-36=0$$

$$-\lambda^2(\lambda-6)+(\lambda-6)(\lambda+6)=0$$

$$-(\lambda-6)(\lambda-3)(\lambda+2)=0$$

The roots are: $\lambda_1=6,\lambda_2=3,\lambda_3=-2.$

These are the eigenvalues.

Find the eigenvectors.

$\lambda_1=6$

$$\begin{bmatrix} 1-\lambda & 1 & 3 \\ 1 & 5-\lambda & 1 \\ 3 & 1 & 1-\lambda \\ \end{bmatrix}=\begin{bmatrix} -5 & 1 & 3 \\ 1 & -1 & 1 \\ 3 & 1 & -5 \\ \end{bmatrix}$$

$R_2=R_2+(1/5)R_1$

$$\begin{bmatrix} -5 & 1 & 3 \\ 0 & -4/5 & 8/5 \\ 3 & 1 & -5 \\ \end{bmatrix}$$

$R_3=R_3+(3/5)R_1$

$$\begin{bmatrix} -5 & 1 & 3 \\ 0 & -4/5 & 8/5 \\ 0 & 8/5 & -16/5 \\ \end{bmatrix}$$

$R_3=R_3+(2)R_2$

$$\begin{bmatrix} -5 & 1 & 3 \\ 0 & -4/5 & 8/5 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

$R_2=(-5/4)R_2$

$$\begin{bmatrix} -5 & 1 & 3 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

$R_1=R_1-R_2$

$$\begin{bmatrix} -5 & 0 & 5 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

$R_1=(-1/5)R_1$

$$\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

Solve the matrix equation

$$\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ \end{bmatrix}\begin{bmatrix} v _1 \\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

If we take $v_3=t,$ then $v_1=t,v_2=2t,v_3=t.$

$$\text{v}=\begin{bmatrix} t \\ 2t \\ t \end{bmatrix}=\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}t$$

$\lambda_2=3$

$$\begin{bmatrix} 1-\lambda & 1 & 3 \\ 1 & 5-\lambda & 1 \\ 3 & 1 & 1-\lambda \\ \end{bmatrix}=\begin{bmatrix} -2 & 1 & 3 \\ 1 & 2 & 1 \\ 3 & 1 & -2 \\ \end{bmatrix}$$

$R_2=R_2+(1/2)R_1$

$$\begin{bmatrix} -2 & 1 & 3 \\ 0 & 5/2 & 5/2 \\ 3 & 1 & -2 \\ \end{bmatrix}$$

$R_3=R_3+(3/2)R_1$

$$\begin{bmatrix} -2 & 1 & 3 \\ 0 & 5/2 & 5/2 \\ 0 & 5/2 & 5/2 \\ \end{bmatrix}$$

$R_3=R_3-R_2$

$$\begin{bmatrix} -2 & 1 & 3 \\ 0 & 5/2 & 5/2 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

$R_2=(2/5)R_2$

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

$R_1=R_1-R_2$

$$\begin{bmatrix} -2 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

$R_1=(-1/2)R_1$

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

Solve the matrix equation

$$\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ \end{bmatrix}\begin{bmatrix} v _1 \\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

If we take $v_3=t,$ then $v_1=t,v_2=-t,v_3=t.$

$$\text{v}=\begin{bmatrix} t \\ -t \\ t \end{bmatrix}=\begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}t$$

$\lambda_3=-2$

$$\begin{bmatrix} 1-\lambda & 1 & 3 \\ 1 & 5-\lambda & 1 \\ 3 & 1 & 1-\lambda \\ \end{bmatrix}=\begin{bmatrix} 3 & 1 & 3 \\ 1 & 7 & 1 \\ 3 & 1 & 3 \\ \end{bmatrix}$$

$R_2=R_2-(1/3)R_1$

$$\begin{bmatrix} 3 & 1 & 3 \\ 0 & 20/3 & 0 \\ 3 & 1 & 3 \\ \end{bmatrix}$$

$R_3=R_3-R_1$

$$\begin{bmatrix} 3 & 1 & 3 \\ 0 & 20/3 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

$R_2=(3/20)R_2$

$$\begin{bmatrix} 3 & 1 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

$R_1=R_1-R_2$

$$\begin{bmatrix} 3 & 0 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

$R_1=(1/3)R_1$

$$\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

Solve the matrix equation

$$\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}\begin{bmatrix} v _1 \\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

If we take $v_3=t,$ then $v_1=-t,v_2=0,v_3=t.$

$$\text{v}=\begin{bmatrix} -t \\ 0 \\ t \end{bmatrix}=\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}t$$

The Normalized matrix

$$N=\begin{bmatrix} 1/\sqrt{6} & 1/\sqrt{3} & -1/\sqrt{2} \\ 2/\sqrt{6} & -1/\sqrt{3} & 0 \\ 1/\sqrt{6} & 1/\sqrt{3} & 1/\sqrt{2} \\ \end{bmatrix}$$

$$N^T=\begin{bmatrix} 1/\sqrt{6} & 2/\sqrt{6} & 1/\sqrt{6} \\ 1/\sqrt{3} & -1/\sqrt{3} & 1/\sqrt{3} \\ -1/\sqrt{2} & 0 & 1/\sqrt{2} \\ \end{bmatrix}$$

$$\begin{bmatrix} 1/\sqrt{6} & 2/\sqrt{6} & 1/\sqrt{6} \\ 1/\sqrt{3} & -1/\sqrt{3} & 1/\sqrt{3} \\ -1/\sqrt{2} & 0 & 1/\sqrt{2} \\ \end{bmatrix}A\begin{bmatrix} 1/\sqrt{6} & 1/\sqrt{3} & -1/\sqrt{2} \\ 2/\sqrt{6} & -1/\sqrt{3} & 0 \\ 1/\sqrt{6} & 1/\sqrt{3} & 1/\sqrt{2} \\ \end{bmatrix}=$$

$$=\begin{bmatrix} 6 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -2 \\ \end{bmatrix}$$

$$\begin{bmatrix} y_1 & y_2 & y_3 \end{bmatrix}\begin{bmatrix} 6 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -2 \\ \end{bmatrix}\begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix}=$$

$$=6y_1^2+3y_2^2-2y_3^2$$

 which is the required canonical form.

$Rank=3$ (the number of non-zero eigenvalues).

$Index=2$ (the number of positive eigenvalues).

$Signature=2-1=1$ (the difference between the number of positive eigenvalues and the number of negative eigenvalues ).

 

Nature: the eigenvalues are $6>0,3>0,-2<0.$ Therefore the given quadratic form is indefinite in nature.