Answer to Question #294734 in Linear Algebra for vaishnavi

Question

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?

Answer 1

$\displaystyle Q(x,y)=2x^2-2xy+2y^2=2x^2-xy-yx+2y^2\\$

$\displaystyle =\begin{pmatrix} x & y \end{pmatrix}\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}$

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

Now, the eigenvalues of $A$ are $\displaystyle \lambda_1=1\text{ and }\lambda_2=3.$

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

$\displaystyle \begin{pmatrix} 1 \\ 1 \end{pmatrix}$

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

$\displaystyle \begin{pmatrix} 1 \\ -1 \end{pmatrix}$

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

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

Using $\displaystyle p_1 \text{ and }p_2$ as column vectors, construct the matrix $P$ such that;

$\displaystyle P=\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix}$

Hence, the transformation $\displaystyle \begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix}\begin{pmatrix} m \\ n \end{pmatrix}$ transforms $Q(x,y)$ to sum of squares or canonical form.

That is;

$\displaystyle Q(x,y)=\begin{pmatrix} x & y \end{pmatrix}\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}$

$\displaystyle \Rightarrow Q(m,n)=\left(\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix}\begin{pmatrix} m \\ n \end{pmatrix}\right)^T\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix}\begin{pmatrix} m \\ n \end{pmatrix}$

$\displaystyle =(m\ \ n)\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix}^T\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix}\begin{pmatrix} m \\ n \end{pmatrix}$

$\displaystyle =(m\ \ n)\begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}\begin{pmatrix} m \\ n \end{pmatrix}$$\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 $Q(x,y)=2$

Index: of a quadratic from is the number of positive square terms in its canonical from. So the index of $Q(x,y)=2$

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