Q ( x , y ) = 2 x 2 − 2 x y + 2 y 2 = 2 x 2 − x y − y x + 2 y 2 \displaystyle
Q(x,y)=2x^2-2xy+2y^2=2x^2-xy-yx+2y^2\\ Q ( x , y ) = 2 x 2 − 2 x y + 2 y 2 = 2 x 2 − x y − y x + 2 y 2
= ( x y ) ( 2 − 1 − 1 2 ) ( x y ) \displaystyle
=\begin{pmatrix}
x & y
\end{pmatrix}\begin{pmatrix}
2 & -1 \\
-1 & 2
\end{pmatrix}\begin{pmatrix}
x \\
y
\end{pmatrix} = ( x y ) ( 2 − 1 − 1 2 ) ( x y )
Since A = ( 2 − 1 − 1 2 ) \displaystyle
A=\begin{pmatrix}
2 & -1 \\
-1 & 2
\end{pmatrix} A = ( 2 − 1 − 1 2 ) is symmetric then it A A A is orthogonally diagonalizable and has real eigenvalues.
Now, the eigenvalues of A A A are λ 1 = 1 and λ 2 = 3. \displaystyle
\lambda_1=1\text{ and }\lambda_2=3. λ 1 = 1 and λ 2 = 3.
An eigenvector corresponding to the eigenvalue λ 1 = 1 \displaystyle
\lambda_1=1 λ 1 = 1 is;
( 1 1 ) \displaystyle
\begin{pmatrix}
1 \\
1
\end{pmatrix} ( 1 1 )
and an eigenvector corresponding to the eigenvalue λ 2 = 3 \displaystyle
\lambda_2=3 λ 2 = 3 is;
( 1 − 1 ) \displaystyle
\begin{pmatrix}
1 \\
-1
\end{pmatrix} ( 1 − 1 )
Hence, the eigenvectors ( 1 , 1 ) and ( 1 , − 1 ) \displaystyle
(1,\ 1)\text{ and }(1,\ -1) ( 1 , 1 ) and ( 1 , − 1 ) form an orthogonal basis for R 2 \R^2 R 2 . Normalizing each of these eigenvectors produces an orthonormal basis;
p 1 = ( 1 , 1 ) ∥ ( 1 , 1 ) ∥ = ( 1 2 , 1 2 ) \displaystyle
p_1=\frac{(1,\ 1)}{\|(1,\ 1)\|}=\left(\frac{1}{\sqrt{2}},\ \frac{1}{\sqrt{2}}\right) p 1 = ∥ ( 1 , 1 ) ∥ ( 1 , 1 ) = ( 2 1 , 2 1 ) , p 2 = ( 1 , − 1 ) ∥ ( 1 , − 1 ) ∥ = ( 1 2 , − 1 2 ) \displaystyle
p_2=\frac{(1,\ -1)}{\|(1,\ -1)\|}=\left(\frac{1}{\sqrt{2}},\ -\frac{1}{\sqrt{2}}\right) p 2 = ∥ ( 1 , − 1 ) ∥ ( 1 , − 1 ) = ( 2 1 , − 2 1 )
Using p 1 and p 2 \displaystyle
p_1 \text{ and }p_2 p 1 and p 2 as column vectors, construct the matrix P P P such that;
P = ( 1 2 1 2 1 2 − 1 2 ) \displaystyle
P=\begin{pmatrix}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}
\end{pmatrix} P = ( 2 1 2 1 2 1 − 2 1 )
Hence, the transformation ( x y ) = ( 1 2 1 2 1 2 − 1 2 ) ( m n ) \displaystyle
\begin{pmatrix}
x \\
y
\end{pmatrix}=\begin{pmatrix}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}
\end{pmatrix}\begin{pmatrix}
m \\
n
\end{pmatrix} ( x y ) = ( 2 1 2 1 2 1 − 2 1 ) ( m n ) transforms Q ( x , y ) Q(x,y) Q ( x , y ) to sum of squares or canonical form.
That is;
Q ( x , y ) = ( x y ) ( 2 − 1 − 1 2 ) ( x y ) \displaystyle
Q(x,y)=\begin{pmatrix}
x & y
\end{pmatrix}\begin{pmatrix}
2 & -1 \\
-1 & 2
\end{pmatrix}\begin{pmatrix}
x \\
y
\end{pmatrix} Q ( x , y ) = ( x y ) ( 2 − 1 − 1 2 ) ( x y )
⇒ Q ( m , n ) = ( ( 1 2 1 2 1 2 − 1 2 ) ( m n ) ) T ( 2 − 1 − 1 2 ) ( 1 2 1 2 1 2 − 1 2 ) ( m n ) \displaystyle
\Rightarrow Q(m,n)=\left(\begin{pmatrix}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}
\end{pmatrix}\begin{pmatrix}
m \\
n
\end{pmatrix}\right)^T\begin{pmatrix}
2 & -1 \\
-1 & 2
\end{pmatrix}\begin{pmatrix}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}
\end{pmatrix}\begin{pmatrix}
m \\
n
\end{pmatrix} ⇒ Q ( m , n ) = ( ( 2 1 2 1 2 1 − 2 1 ) ( m n ) ) T ( 2 − 1 − 1 2 ) ( 2 1 2 1 2 1 − 2 1 ) ( m n )
= ( m n ) ( 1 2 1 2 1 2 − 1 2 ) T ( 2 − 1 − 1 2 ) ( 1 2 1 2 1 2 − 1 2 ) ( m n ) \displaystyle
=(m\ \ n)\begin{pmatrix}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}
\end{pmatrix}^T\begin{pmatrix}
2 & -1 \\
-1 & 2
\end{pmatrix}\begin{pmatrix}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}
\end{pmatrix}\begin{pmatrix}
m \\
n
\end{pmatrix} = ( m n ) ( 2 1 2 1 2 1 − 2 1 ) T ( 2 − 1 − 1 2 ) ( 2 1 2 1 2 1 − 2 1 ) ( m n )
= ( m n ) ( 1 0 0 3 ) ( m n ) \displaystyle
=(m\ \ n)\begin{pmatrix}
1 & 0 \\
0 & 3
\end{pmatrix}\begin{pmatrix}
m \\
n
\end{pmatrix} = ( m n ) ( 1 0 0 3 ) ( m n ) = m 2 + 3 n 2 \displaystyle
=m^2+3n^2 = m 2 + 3 n 2
which is the canonical form of the given quadratic form.
Nature: it is positive definite in nature since all eigenvalues are positive numbers.
Rank: of a quadratic from is the number of square terms in its canonical form. So rank of Q ( x , y ) = 2 Q(x,y)=2 Q ( x , y ) = 2
Index: of a quadratic from is the number of positive square terms in its canonical from. So the index of Q ( x , y ) = 2 Q(x,y)=2 Q ( x , y ) = 2
Signature: of Q ( x , y ) = Q(x,y)= Q ( x , y ) = (number of positive square terms)- (number of negative square terms)= 2 =2 = 2
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