Answer to Question #102077 in Linear Algebra for Gayatri Yadav

Question

Find the orthogonal canonical reduction of the quadratic form
−

Answer 1

Find the orthogonal canonical reduction of the quadratic form

$$Q=x^2+y^2+z^2-2xy-2xz-2yz$$

The quadratic form can be expressed in matrix notation as

$$Q=\bold x^TA\bold x=\begin{pmatrix} x & y & z \end{pmatrix}^T\begin{pmatrix} 1 & -1 & -1 \\ -1 & 1 & -1 \\ -1 & -1 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}$$

The characteristic equation of the matrix $A$ is

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

$(1-\lambda)\begin{vmatrix} 1-\lambda & -1 \\ -1 & 1-\lambda \end{vmatrix}-(-1)\begin{vmatrix} -1 & -1 \\ -1 & 1-\lambda \end{vmatrix}+$$(-1)\begin{vmatrix} -1 & 1-\lambda \\ -1& -1 \end{vmatrix}=0$

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

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

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

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

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

$\lambda_1=\lambda_2=2, \lambda_3=-1$ are eigenvalues.

$Q=2x'^2+2y'^2-z'^2$ is the orthogonal canonical reduction.

$\lambda=2: \begin{pmatrix} 1-\lambda & -1 & -1 \\ -1 & 1-\lambda & -1 \\ -1 & -1 & 1-\lambda \end{pmatrix}=\begin{pmatrix} -1 & -1 & -1 \\ -1 & -1 & -1 \\ -1 & -1 & -1 \end{pmatrix}$

Perform row operations to obtain the rref of the matrix:

$R_2=R_2-R_1: \begin{pmatrix} -1 & -1 & -1 \\ 0 & 0 & 0 \\ -1 & -1 & -1 \end{pmatrix}$

$R_3=R_3-R_1: \begin{pmatrix} -1 & -1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$

$R_1=-1\cdot R_1: \begin{pmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$

Now, solve the matrix equation

$$\begin{pmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$

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

Therefore

$$\bold v=\begin{pmatrix} -t-s \\ t \\ s \end{pmatrix}=\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}t+\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}s$$

$\lambda=-1: \begin{pmatrix} 1-\lambda & -1 & -1 \\ -1 & 1-\lambda & -1 \\ -1 & -1 & 1-\lambda \end{pmatrix}=\begin{pmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{pmatrix}$

Perform row operations to obtain the rref of the matrix:

$R_1=R_1/2: \begin{pmatrix} 1 & -1/2 & -1/2 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{pmatrix}$

$R_2=R_2+R_1: \begin{pmatrix} 1 & -1/2 & -1/2 \\ 0 & 3/2 & -3/2 \\ -1 & -1 & 2 \end{pmatrix}$

$R_3=R_3+R_1: \begin{pmatrix} 1 & -1/2 & -1/2 \\ 0 & 3/2 & -3/2 \\ 0 & -3/2 & 3/2 \end{pmatrix}$

$R_3=R_3+R_2: \begin{pmatrix} 1 & -1/2 & -1/2 \\ 0 & 3/2 & -3/2 \\ 0 & 0 & 0 \end{pmatrix}$

$R_2=(2/3)R_2: \begin{pmatrix} 1 & -1/2 & -1/2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix}$

$R_1=R_1+(1/2)R_2: \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix}$

Now, solve the matrix equation

$$\begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$

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

$$\bold v=\begin{pmatrix} t \\ t \\ t \end{pmatrix}=\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}t$$

$\sqrt {(-1)^2+1^2+0^2}=\sqrt{2}$

$\sqrt {(-1)^2+0^2+1^2}=\sqrt{2}$

$\sqrt {1^2+1^2+1^2}=\sqrt{3}$

$p_1={1 \over \sqrt{2}}\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix},p_2={1 \over \sqrt{2}}\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix},p_3={1 \over \sqrt{3}}\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ are the principal axes.

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

Thus, a substitution $\bold x=P\bold x'$ that eliminates the cross product terms is

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

This produces the new quadratic form

$Q=\bold x'^T(P^TAP)\bold x'=$

$=\begin{pmatrix} x' & y' & z' \end{pmatrix}^T\begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -1 \end{pmatrix}\begin{pmatrix} x' \\ y' \\ z' \end{pmatrix}=$

$=2x'^2+2y'^2-z'^2$

in which there are no cross product terms.

$Q=2x'^2+2y'^2-z'^2$ is the orthogonal canonical reduction.