Answer to Question #106169 in Linear Algebra for Sourav mondal

Question

Reduce the quadratic form 8x²— 4xy + 5y² to
its normal canonical form.

Answer 1

Reduce $8x^2-4xy+5y^2$ to its normal canonical form.

It is possible to solve this problem using orthogonal transformation.

Let it be a matrix $A$ such that

$A = \begin{bmatrix} \mathrm{coef} x_1^2 & \frac{1}{2} \mathrm{coef} x_1x_2 \\ \frac{1}{2}\mathrm{coef} x_2x_1 & \mathrm{coef} x_2^2 \end{bmatrix}$

Remembering that $x_1x_2=x_2x_1$ it is possible to calculate $A$:

$A = \begin{bmatrix} 8 & -2 \\ -2 & 5 \end{bmatrix}$

To express the given quadratic form in canonical form it is necessary to diagonalize matrix $A$, that can be done by the following formula:

$M^{-1}AM =D$,

where $M$ is normilized eigenvectors matrix and $D$ is diagonalized matrix $A$.

To find $M$ it is necessary to find eigenvalues and eigenvectors of matrix $A$:

$\det(\lambda I-A)=0 \Leftrightarrow \begin{vmatrix} \lambda-8 & 2 \\ 2 & \lambda-5 \end{vmatrix} = (\lambda-8)(\lambda-5)-4=\lambda^2-13+36=0$

This equation can be solved with usage of Vieta's formulas:

$\begin{cases} \lambda_1+\lambda_2=13 \\ \lambda_1 \cdot \lambda_2 = 36 \end{cases} \Bigg| \Rightarrow \begin{cases} \lambda_1=4 \\ \lambda_2=9 \end{cases}$

Eigenvector, corresponding to eigenvalue $\lambda_i$ can be found with solving the following equation:

$Av_i=\lambda_iv_i$,

where $v_i$ is eigenvector.

1. Eigenvector, corresponding to eigenvalue $\lambda_1=4$:

$Av_1=\lambda_iv_1 \Leftrightarrow \begin{cases} 8v_{11}-2v_{12}=4v_{11}\\ -2v_{11}+5v_{12}=4v_{12} \end{cases} \Bigg| \Leftrightarrow \begin{cases} v_{12}=2v_{11}\\ -2v_{11}+10v_{11}=8v_{11} \end{cases} \Bigg| \Leftrightarrow \begin{cases} v_{12}=2v_{11}\\ 8v_{11}=8v_{11} \end{cases} \Bigg| \Leftrightarrow \begin{cases} v_{12}=2v_{11}\\ v_{11} \mathrm{-any\ number} \end{cases}$

Let $v_{11}=1$ , so $v_{12} = 2 v_{11} = 2$, so normilized vector $v_1$ can be found as:

$\overline{v}_1 = \begin{bmatrix} \frac{v_{11}}{\sqrt{v_{11}^2+v_{12}^2}} \\ \frac{v_{12}}{\sqrt{v_{11}^2+v_{12}^2}} \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{1^2+2^2}} \\ \frac{2}{\sqrt{1^2+2^2}} \end{bmatrix} =\begin{bmatrix} \frac{1}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} \end{bmatrix}$

2. Eigenvector, corresponding to eigenvalue $\lambda_2=9$:

$Av_2=\lambda_iv_2 \Leftrightarrow \begin{cases} 8v_{21}-2v_{22}=9v_{21}\\ -2v_{21}+5v_{22}=9v_{12} \end{cases} \Bigg| \Leftrightarrow \begin{cases} v_{22}=-\frac{1}{2}v_{21}\\ -2v_{21}-\frac{5}{2}v_{21}=-\frac{9}{2}v_{21} \end{cases} \Bigg| \Leftrightarrow \begin{cases} v_{22}=-\frac{1}{2}v_{21}\\ -\frac{9}{2}v_{21}=-\frac{9}{2}v_{21} \end{cases} \Bigg| \Leftrightarrow \begin{cases} v_{22}=-\frac{1}{2}v_{21}\\ v_{21} \mathrm{-any\ number} \end{cases}$

Let $v_{21}=1$ , so $v_{22} = -\frac{1}{2}v_{21} = -\frac{1}{2}$, so normilized vector $v_2$ can be found as:

$\overline{v}_2 = \begin{bmatrix} \frac{v_{21}}{\sqrt{v_{21}^2+v_{22}^2}} \\ \frac{v_{22}}{\sqrt{v_{21}^2+v_{22}^2}} \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{1^2+(-\frac{1}{2})^2}} \\ \frac{-\frac{1}{2}}{\sqrt{1^2+(-\frac{1}{2})^2}} \end{bmatrix} =\begin{bmatrix} \frac{2}{\sqrt{5}} \\ -\frac{1}{\sqrt{5}} \end{bmatrix}$

Combining $\overline{v}_1$ and $\overline{v}_2$ it is possible to get $M$:

$M = \begin{bmatrix} \overline{v}_1 & \overline{v}_2 \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} & -\frac{1}{\sqrt{5}} \end{bmatrix} = \frac{1}{\sqrt{5}} \cdot \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix}$

Inverse matrix $M^{-1}$ can be found as:

$M^{-1} = \cfrac{1}{\det M} \cdot \cfrac{1}{\sqrt{5}} \begin{bmatrix} -1 & -2 \\ -2 & 1 \end{bmatrix}^T = \cfrac{1}{\det M} \cdot \cfrac{1}{\sqrt{5}} \begin{bmatrix} -1 & -2 \\ -2 & 1 \end{bmatrix}$

Determinant of matrix $M$ equals to:

$\det M = \cfrac{1}{\sqrt{5}} \cdot (-\cfrac{1}{\sqrt{5}}) - \cfrac{2}{\sqrt{5}} \cdot \cfrac{2}{\sqrt{5}} = -\cfrac{1}{5} - \cfrac{4}{5}=-1$

So, inverse matrix $M^{-1}$ is equal to:

$M^{-1} = \cfrac{1}{\det M} \cdot \cfrac{1}{\sqrt{5}} \begin{bmatrix} -1 & -2 \\ -2 & 1 \end{bmatrix} = \cfrac{1}{-1} \cdot \cfrac{1}{\sqrt{5}} \begin{bmatrix} -1 & -2 \\ -2 & 1 \end{bmatrix} = \cfrac{1}{\sqrt{5}} \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix}$

So, matrix $D$ equals to:

$D = M^{-1} A M = \cfrac{1}{\sqrt{5}} \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} \cdot \begin{bmatrix} 8 & -2 \\ -2 & 5 \end{bmatrix} \cdot \cfrac{1}{\sqrt{5}} \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} = \cfrac{1}{5} \cdot \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} \cdot \begin{bmatrix} 8 & -2 \\ -2 & 5 \end{bmatrix} \cdot \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} = \cfrac{1}{5} \begin{bmatrix} 4 & 8 \\ 18 & -9 \end{bmatrix} \cdot \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} = \cfrac{1}{5} \cdot \begin{bmatrix} 20 & 0 \\ 0 & 45 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 9 \end{bmatrix}$

So, the canonical form can be found as:

$D_{11} \hat{x}^2 + D_{22} \hat{y}^2 =4\hat{x}^2+9\hat{y}^2$

Answer: $4\hat{x}^2+9\hat{y}^2$