2.Eliminate Cross-product term

A quadratic form may be rewritten as the sum of two parts :

$q(\rm\bold x)=\rm\bold x^{\it T}\it A \rm\bold x =\it \sum _{i=1}^n a_{ii}x_{i}^2+ \rm2\it \sum_{i

in which the frist part   $\dpi{100} \it \sum_{i=1}^{n} a_{ii}x_{i}^2$ is called the (perfect) square terms and the second part  $\it \sum_{i is called the cross-product terms(교차곱). Actually, what makes it hard to sketch the quadratic surface is the cross product terms. However, the symmetric matrix $\dpi{100} A$ can be orthogonally diagonalized(직교 대각화), i,e., there exists an orthogonal matrix  $\dpi{100} P$ such that

$\dpi{100} P^{T}AP=P^{-1}AP=D=\begin{bmatrix} \lambda_1 & & &0 \\ &\lambda_2 & & \\ & &\ddots & \\ 0 & & &\lambda_n \end{bmatrix}$ .

Here, the diagonal entries  $\dpi{100} \lambda_i$'s are the eigenvalues of $\dpi{100} A$ and the column vectors of $\dpi{100} P$ are their associated eigenvectors of $\dpi{100} A$. Then we get, by setting  $\dpi{100} \rm\bold x =\it P\rm\bold y$,

$\dpi{100} \rm\bold x^{\it T}\it A\rm\bold x=\rm\bold y^{\it T}\it (P^{T}AP)\rm\bold y=\rm\bold y^{\it T}\it D\rm\bold y =\it \lambda_1 y_1^2+\lambda_2 y_2^2+\cdots+\lambda_n y_n^2$,

which is a quadratic form without the cross-product terms. Consequently, we have proven the following theorem.

Theorem 2 (Principal axis Theorem: 주축정리)

Let  $\dpi{100} \rm\bold x^{\it T}\it A\rm\bold x$ be a quadratic form in  $\dpi{100} \rm\bold x=\it [x_1, x_2, \cdots, x_n]^{T} \in \mathbb{R}^n$  for a symmetric matrix $\dpi{100} A$. then there is a change of coordinate of $\dpi{100} \rm\bold x$ into  $\dpi{100} \rm\bold y=\it P^{T}\rm\bold x =\it [y_1,y_2,\cdots,y_n]^T$ such that  $\dpi{100} P^TAP=D$.

$\dpi{100} \rm\bold x^{\it T}\it A\rm\bold x=\rm\bold y^{\it T}\it D\rm\bold y =\it \lambda_1y_1^2+\lambda_2y_2^2+\cdots+\lambda_n^2$ ,

where $\dpi{100} P$ is an orthogonal matrix(직교행렬) and $\dpi{100} P^TAP=D$

Example)  Eliminate the cross-product terms in the conic section  $3x^2+2xy+3y^2-8=0$.

This equation can be written in the form

The matrix  $A=\begin{bmatrix} 3 &1 \\ 1 & 3 \end{bmatrix}$  has eigenvalues $\lambda_1 =2, \lambda_2=4$ with associated eigenvectors  $\rm\bold v_1=(1,-1)$  and $\rm\bold v_2=(1,1)$

the unit vectors  $\rm\bold u_1=(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$  and $\rm\bold u_2=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$.

Then change coordinates, $\rm\bold x= \it P\rm\bold y$   for  $P=\frac{1}{\sqrt{2}}\begin{bmatrix} 1 &1 \\ -1&1 \end{bmatrix}$  . i,e,

다음은 이변수 이차형식에 대해 주축정리에 의해 교차곱을 없앤 이차형식의 형태와 그 행렬을 나타내주는 interact입니다.

Example)  Eliminate the cross-product terms in the conic section  $2xy+2xz-1=0$.

The matrix for the given quadratic form is

The eigenvalues  of $A$ are  $\lambda_1=\sqrt{2},\lambda_2=-\sqrt{2}, \lambda_3=0$ and associated eigenvectors are  $\rm\bold v_1=(1,-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}),\rm\bold v_2=(1,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}),\rm\bold v_3=(0,1,-1)$ .

Hence, orthogonal matrix $P$ that diagonalizes $A$ is

Thus, the equation is transformed to $\sqrt{2}x1^2-\sqrt{2}y2^2=1$, which is a hyperbolic cylinder.

다음은 3변수에 이차형식에 대해 주축정리에 의해 교차곱이 없어진 형태와 그 행렬을 나타내주는 interact입니다.

좀더 자세한 내용 (참조: http://math1.skku.ac.kr/home/pub/751/ 2013.11.27)

http://matrix.skku.ac.kr/2013-Album/SKKU-LA-SolutionsSPF/Image546.jpg2013.11.27

3. 선형대수학 Section 8-4, Quadratic Ft

http://youtu.be/lznsULrqJ_0 2013.11.27

Reference:

Diagonalization of quadratic form, Theorem5 & Examples : Linear Algebra, Kwark&Hong, Birkhauser, P282~p285

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