Answer to Question #294734 in Linear Algebra for vaishnavi

Question #294734

Find the conical form of a quadratic form Q(x,y)=2x^2+2y^2-2xy by using an orthogonal transformation hence find nature,rank,index and signature of the conical form?

1
Expert's answer
2022-02-08T12:32:19-0500

Q(x,y)=2x22xy+2y2=2x2xyyx+2y2\displaystyle Q(x,y)=2x^2-2xy+2y^2=2x^2-xy-yx+2y^2\\

=(xy)(2112)(xy)\displaystyle =\begin{pmatrix} x & y \end{pmatrix}\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}

Since A=(2112)\displaystyle A=\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix} is symmetric then it AA is orthogonally diagonalizable and has real eigenvalues.

Now, the eigenvalues of AA are λ1=1 and λ2=3.\displaystyle \lambda_1=1\text{ and }\lambda_2=3.


An eigenvector corresponding to the eigenvalue λ1=1\displaystyle \lambda_1=1 is;

(11)\displaystyle \begin{pmatrix} 1 \\ 1 \end{pmatrix}

and an eigenvector corresponding to the eigenvalue λ2=3\displaystyle \lambda_2=3 is;

(11)\displaystyle \begin{pmatrix} 1 \\ -1 \end{pmatrix}

Hence, the eigenvectors (1, 1) and (1, 1)\displaystyle (1,\ 1)\text{ and }(1,\ -1) form an orthogonal basis for R2\R^2. Normalizing each of these eigenvectors produces an orthonormal basis;

p1=(1, 1)(1, 1)=(12, 12)\displaystyle p_1=\frac{(1,\ 1)}{\|(1,\ 1)\|}=\left(\frac{1}{\sqrt{2}},\ \frac{1}{\sqrt{2}}\right), p2=(1, 1)(1, 1)=(12, 12)\displaystyle p_2=\frac{(1,\ -1)}{\|(1,\ -1)\|}=\left(\frac{1}{\sqrt{2}},\ -\frac{1}{\sqrt{2}}\right)


Using p1 and p2\displaystyle p_1 \text{ and }p_2 as column vectors, construct the matrix PP such that;

P=(12121212)\displaystyle P=\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix}

Hence, the transformation (xy)=(12121212)(mn)\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} transforms Q(x,y)Q(x,y) to sum of squares or canonical form.


That is;

Q(x,y)=(xy)(2112)(xy)\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(m,n)=((12121212)(mn))T(2112)(12121212)(mn)\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}

=(m  n)(12121212)T(2112)(12121212)(mn)\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)(1003)(mn)\displaystyle =(m\ \ n)\begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}\begin{pmatrix} m \\ n \end{pmatrix}=m2+3n2\displaystyle =m^2+3n^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)=2Q(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)=2Q(x,y)=2

Signature: of Q(x,y)=Q(x,y)= (number of positive square terms)- (number of negative square terms)=2=2


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