CSIR NET MATHEMATICS 2020-JUNE

HELD ON 26TH-NOVEMBER

PART-B

5.Which of the following real quadratic forms on $R^2$ is positive definite?

(A)$Q(X,Y)=XY$

(B)$Q(X,Y)=(X^2+XY)$

(C)$Q(X,Y)=X^2+2XY+Y^2$

(D)$Q(X,Y)=X^2-XY+Y^2$

Solution:Let quadratic form on $R^2=(X,Y)$

let denote $Q(X,Y)=aX^2+bXY+cY^2$ is a quadratic form on $R^2$

The matrx corresponding to the quadratic form is

Q(X,Y)=$ \begin{bmatrix}
co.efficient of X^2 & \frac{1}{2} co.efficient of XY \\
\frac{1}{2} co.efficient of YX & co.efficient of Y^2
\end{bmatrix} $

Q(X,Y) =$ \begin{bmatrix}
   a & \frac{1}{2} b \\
   \frac{1}{2} b &     c
\end{bmatrix}$

Test the Nature of a quadratic form through principal minors

Let $A=[a_{ij}]$ be the matrix of the quadratic form in n variables $x_1,x_2,\dots,x_n$.Then A is a square symmetric matrix of order n

Let $D_1,D_2,D_3,\dots,D_n $are the principal minors of A

[i] The Qudratic form is Positive definite if $D_1,D_2,D_3,\dots,D_n $ are all positive
[ i.e D_n>0 forall n ]

[ii] The Qudratic form is Negative definite if $D_1,D_3,D_5,\dots, $ are all negative e and $D_2,D_4,D_6,\dots, $ are all positive
[ i.e, (-1)^n D_n >0 forall n ]

[iii] The Qudratic form is Positive semi definite if $D_n \geq 0$ are all positive and atleast one $D_i=0$

[iv] The Qudratic form is Negative semi definite if $(-1)^n D_n \geq 0$ and atleast one $D_i=0$

[v] The Qudratic form is indefinite in all other cases

OPTION A: $Q(X,Y)=XY$

The matrix corresponding to the quadratic form is

    Q(X,Y) = $ \begin{bmatrix}
        0 & \frac{1}{2} \\
        \frac{1}{2} & 0
   \end{bmatrix} $

Let the principal minors of $Q(X,Y)$

$ \vert D_1 \vert =\begin{vmatrix}
0
\end{vmatrix} $=0

=$ \begin{vmatrix}
0
\end{vmatrix} $ =0

Hence for all $D_1 \geq 0$

    $ \vert D_2 \vert =\begin{vmatrix}
            0 & \frac{1}{2} \\
        \frac{1}{2} & 0
   \end{vmatrix}$ = $ 0-\frac{1}{2} \times \frac{1}{2} $=$ 0-\frac{1}{4} $

=-1/4

Hence $D_2 < 0 $

$ and $ D_2 <0 $ is indefinte.

So option A is not correct

OPTION B $Q(X,Y)=(X^2+XY)$

The matrix corresponding to the quadratic form is

    Q(X,Y)=$ \begin{bmatrix}
        1 & \frac{1}{2} \\
        \frac{1}{2} & 0
   \end{bmatrix} $

Let the principal minors of $Q(X,Y)$

$ \vert D_1 \vert =\begin{vmatrix}
        1
   \end{vmatrix}$ =1

=$\begin{vmatrix}
0
\end{vmatrix} $=0

Hence for all $D_1 \geq 0$

   $ \vert D_2 \vert =\begin{vmatrix}
        1 & \frac{1}{2} \\
        \frac{1}{2} & 0
   \end{vmatrix}=0-\frac{1}{2} \times\frac{1}{2}=0-\frac{1}{4}=\frac{-1}{4} $

Hence $D_2 < 0 $

i.e., $D_1\geq 0 $ and $ D_2 <0 $ is indefinite

So option B is not correct

OPTION C $Q(X,Y)=(X^2+2XY+Y^2)$

The matrix corresponding to the quadratic form is

Q(X,Y)=$ \begin{bmatrix}
        1 & \frac{1}{2}2 \\
        \frac{1}{2} 2 & 1
   \end{bmatrix}$ =$ \begin{bmatrix}
   1 & 1 \\
1 & 1
\end{bmatrix} $

Let the principal minors of $Q(X,Y)$

$\vert D_1 \vert =\begin{vmatrix}
1
\end{vmatrix}$ =1

   =$ \begin{vmatrix}
        1
   \end{vmatrix} $ =1

Hence for all $D_1 > 0$

   $ \vert D_2 \vert =\begin{vmatrix}
        1 & 1 \\
        1 & 1
   \end{vmatrix} $=1-1 =0

Hence $D_2 =0 $

i.e., $D_1 > 0 $ and $ D_2 =0 $ is Positive semi definite

So option C is not correct

OPTION D $Q(X,Y)=X^2-XY+Y^$2

The matrix corresponding to the quadratic form is

    Q(X,Y)=$ \begin{bmatrix}
        1 & \frac{1}{2}(-1) \\
        \frac{1}{2} (-1) & 1
   \end{bmatrix}$ =$ \begin{bmatrix}
        1 & \frac{-1}{2} \\
        \frac{-1}{2} & 1
   \end{bmatrix} $

Let the principal minors of $Q(X,Y)$

    $ \vert D_1 \vert =\begin{vmatrix}
        1
   \end{vmatrix} $ =1

    =$ \begin{vmatrix}
        1
   \end{vmatrix} $ =1

Hence for all $D_1 > 0$

   $ \vert D_2 \vert =\begin{vmatrix}
        1 & \frac{-1}{2} \\
   \frac{-1}{2} & 1
   \end{vmatrix} $ =$ 1-\frac{-1}{2}\times \frac{-1}{2} =1-\frac{1}{4}=\frac{4-1}{4}=\frac{3}{4} $

Hence $D_2 >0 $

i.e., $D_1 > 0 $ and $ D_2 >0 $ is Positive definite

So option (D) $Q(X,Y)=X^2-XY+Y^$2 is correct