Question #288172

Reduce the quadratic form X2+3Y2+3Z2-2yz to the canonical form through orthogonal transformation

1
Expert's answer
2022-01-18T07:16:56-0500

We shall reduce the quadratic form x2+3y2+3z22yzx^{2}+3y^{2}+3z^{2}-2yz to the canonical form through the orthogonal transformation.


NOTE: WE SHALL RE-REPRESENT THE VARAIBALE x, y and z TO VARIABLES x1,x2andx3x_{1},x_{2}\hspace{0.1cm}and\hspace{0.1cm}x_{3} respectively.


NOW, using x1,x2andx3x_{1},x_{2}\hspace{0.1cm}and\hspace{0.1cm}x_{3}

Q(x)=x12+3x22+3x322x2x3\displaystyle Q(x)=x_{1}^{2}+3x_{2}^{2}+3x_{3}^{2}-2x_{2}x_{3}

The symmetric matrix A with the given quadratic form is

A=A= [100031013]\begin{bmatrix} 1 & 0 & 0 \\ 0 & 3 & -1\\ 0 & -1 & 3 \end{bmatrix}

The characteristic equation of A is

(AλI)=λ37λ2+14λ8=0    λ=1,2,4\mid(A-\lambda I)\mid=\lambda^{3}-7\lambda^{2}+14\lambda-8=0\\\implies \lambda=1,2,4


The corresponding eigen values are:

for λ1=1,isX1=k[100]\lambda_{1}=1,\hspace{0.2cm}is\hspace{0.2cm}X_{1}=k\begin{bmatrix} 1 \\ 0\\ 0 \end{bmatrix}


for λ2=2,isX2=k[011]\lambda_{2}=2,\hspace{0.2cm}is\hspace{0.2cm}X_{2}=k\begin{bmatrix} 0 \\ 1\\ 1 \end{bmatrix}


for λ3=1,isX3=k[011]\lambda_{3}=1,\hspace{0.2cm}is\hspace{0.2cm}X_{3}=k\begin{bmatrix} 0 \\ 1\\ -1 \end{bmatrix}


Clearly, the eigen vectors X1=[100]X_{1}=\begin{bmatrix} 1 \\ 0\\ 0 \end{bmatrix} , X2=[011]X_{2}=\begin{bmatrix} 0 \\ 1\\ 1 \end{bmatrix} and X3=[011]X_{3}=\begin{bmatrix} 0 \\ 1\\ -1 \end{bmatrix} are linearly independent and orthogonal.


The normalized vectors of the above eigen vextors X1,X2,X3X_{1}, X_{2},X_{3} are:

e1=X1X1=[100]e_{1}=\frac{X_{1}}{\mid\mid X_{1}\mid\mid}= \begin{bmatrix} 1 \\ 0\\ 0 \end{bmatrix} , e2=X2X2=[01212]\displaystyle e_{2}=\frac{X_{2}}{\mid\mid X_{2}\mid\mid}= \begin{bmatrix} 0 \\ \frac{1}{\sqrt2}\\ \frac{1}{\sqrt2} \end{bmatrix} , e3=X3X3=[01212]\displaystyle e_{3}=\frac{X_{3}}{\mid\mid X_{3}\mid\mid}= \begin{bmatrix} 0 \\ \frac{1}{\sqrt2}\\ -\frac{1}{\sqrt2} \end{bmatrix}


Therefore,

The normalized model matrix is

pˉ=[1000121201212]\displaystyle \bar{p}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}} \\ 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix} and pˉ1=pˉT=12[200011011]\bar{p}^{-1}=\bar{p}^{T}=\frac{1}{\sqrt{2}}\displaystyle \begin{bmatrix} \sqrt{2} & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & - 1 \end{bmatrix}


pˉ1Apˉ=12[200011011][ 100031013][1000121201212]\bar{p}^{-1}A\bar{p}= \frac{1}{\sqrt{2}}\displaystyle \begin{bmatrix} \sqrt{2} & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & - 1 \end{bmatrix}\begin{bmatrix} \ 1 & 0 & 0 \\ 0 & 3 & -1 \\ 0 & -1 & 3 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}} \\ 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}


=[ 100011011]=D=\begin{bmatrix} \ 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & -1 \end{bmatrix}=D


We know that the quadratic form is Q=XTAXQ=X^{T}AX where X=PˉYX=\bar{P}Y

then,

Q=(PˉY)A(PˉY)Q=(\bar{P}Y)A(\bar{P}Y)

=(YTPˉT)A(PˉY)=YT(PTAPˉ)Y=YT(Pˉ1APˉ)YsincePˉT=Pˉ1=YTDY=(Y^{T}\bar{P}^{T})A(\bar{P}Y)\\=Y^{T}(P^{T}A\bar{P})Y\\=Y^{T}(\bar{P}^{-1}A\bar{P})Y\hspace{0.3cmsince\hspace{0.3cm}}\bar{P}^{T}=\bar{P}^{-1}\\=Y^{T}DY

Therefore, the canonical form

=[y1y2y3][ 100020004][ y1y2y3]=\begin{bmatrix} y_{1} & y_{2} &y_{3} \\ \end{bmatrix} \begin{bmatrix} \ 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{bmatrix} \begin{bmatrix} \ y_{1} \\ y_{2} \\ y_{3} \end{bmatrix}


=y12+2y22+4y32= y_{1}^{2}+2y_{2}^{2}+4y_{3}^{2}

which gives the canonical form.

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