Answer to Question #248558 in Linear Algebra for chaitu

Question

26. Reduce the Quadratic Form (Q.F.) 2x

2 + 5y

2 + 3z

2 + 4xy to canonical form by an

orthogonal transformation. Also find its nature, rank, index and signature of the Q.F.

Answer 1

The given quadratic form is

$3x^2+5y^2+3z^2-2xy-2yz+2xz$

The matrix of the given quadratic form is

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

We write , $A= IAI$

i,e, $\begin {pmatrix} 3&-1&1 \\ -1&5&-1 \\ 1&-1&3 \end{pmatrix}$ $=\begin{pmatrix} 1&0&0 \\ 0&1&0\\ 0&0&1 \end{pmatrix}$ $A\begin{pmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{pmatrix}$

Now we shall reduce $A$ to diagonal form by applying congruence operation on it . Performing $R_2\rightarrow 3R_2+R_1,R_3\rightarrow 3R_3-R_1;C_2\rightarrow C_2+\frac{1}{3}C_1,$

$C_3\rightarrow C_3-\frac{1}{3}C_1;R_3\rightarrow 7R_3+R_2;C_3\rightarrow C_3-\frac{1}{7}C_2$

We get ,

$\begin{pmatrix} 3&0&0\\ 0&14&0\\ 0&0&54 \end{pmatrix}$ $=$ $\begin{pmatrix} 1&0&0 \\ 1&3&0\\ -6&3&21 \end{pmatrix}$ $A\begin{pmatrix} 1& \frac{1}{3} & \frac{-8}{21} \\ 0&1&\frac{-1}{7} \\ 0&0&1 \end{pmatrix}$

Performing , $R_1\rightarrow \frac{1}{\sqrt{3}}R_1, C_1\rightarrow \frac{1}{\sqrt{3}}C_1$ ;

$R_2\rightarrow \frac{1}{\sqrt{14}}R_1,C_2\rightarrow \frac{1} {\sqrt{14}};$

$R_3\rightarrow \frac{1}{\sqrt{54}}R_3,C_3\rightarrow \frac{1}{\sqrt{54}}C_2$

We get ,

$\begin{pmatrix} 1&0&0 \\ 0&1&0\\ 0&0&1 \end{pmatrix}$ $=\begin{pmatrix} a&0&0\\ b&3b&0\\ -6c&3c&21c \end{pmatrix}A$ $\begin{pmatrix} a&\frac{b}{3}&\frac{-8c}{21} \\ 0&b&\frac{-c}{7} \\ 0&0&c \end{pmatrix}$

Where $a=\frac{1}{\sqrt{3}},b=\frac{1}{\sqrt{14}},c=\frac{1}{\sqrt{54}}$

Thus the linear transformation ,

$X=PY$ where ,

$P=\begin{pmatrix} a&\frac{b}{3}&\frac{-8c}{21} \\ 0&b&\frac{-c}{7} \\ 0&0&c \end{pmatrix}$

$X=[x,y,z]',Y=[y_1,y_2,y_3]'$

This transformation gives quadratic form to the normal form

${y_1}^2+{y_2}^2+{y_3}^2.............(1)$

The rank $r$ of the given quadratic form $=$ The number of nonzero terms in its normal form $(1)$

$=$ 3.

The signature of the given quadratic form $=$ the excess of the number of positive terms over the number of negative terms in its normal form$=3-0=3$