We shall reduce the quadratic form x 2 + 3 y 2 + 3 z 2 − 2 y z x^{2}+3y^{2}+3z^{2}-2yz x 2 + 3 y 2 + 3 z 2 − 2 yz to the canonical form through the orthogonal transformation.
NOTE: WE SHALL RE-REPRESENT THE VARAIBALE x, y and z TO VARIABLES x 1 , x 2 a n d x 3 x_{1},x_{2}\hspace{0.1cm}and\hspace{0.1cm}x_{3} x 1 , x 2 an d x 3 respectively.
NOW, using x 1 , x 2 a n d x 3 x_{1},x_{2}\hspace{0.1cm}and\hspace{0.1cm}x_{3} x 1 , x 2 an d x 3
Q ( x ) = x 1 2 + 3 x 2 2 + 3 x 3 2 − 2 x 2 x 3 \displaystyle Q(x)=x_{1}^{2}+3x_{2}^{2}+3x_{3}^{2}-2x_{2}x_{3} Q ( x ) = x 1 2 + 3 x 2 2 + 3 x 3 2 − 2 x 2 x 3
The symmetric matrix A with the given quadratic form is
A = A= A = [ 1 0 0 0 3 − 1 0 − 1 3 ] \begin{bmatrix}
1 & 0 & 0 \\
0 & 3 & -1\\
0 & -1 & 3
\end{bmatrix} ⎣ ⎡ 1 0 0 0 3 − 1 0 − 1 3 ⎦ ⎤
The characteristic equation of A is
∣ ( A − λ I ) ∣ = λ 3 − 7 λ 2 + 14 λ − 8 = 0 ⟹ λ = 1 , 2 , 4 \mid(A-\lambda I)\mid=\lambda^{3}-7\lambda^{2}+14\lambda-8=0\\\implies \lambda=1,2,4 ∣ ( A − λ I ) ∣= λ 3 − 7 λ 2 + 14 λ − 8 = 0 ⟹ λ = 1 , 2 , 4
The corresponding eigen values are:
for λ 1 = 1 , i s X 1 = k [ 1 0 0 ] \lambda_{1}=1,\hspace{0.2cm}is\hspace{0.2cm}X_{1}=k\begin{bmatrix}
1 \\
0\\
0
\end{bmatrix} λ 1 = 1 , i s X 1 = k ⎣ ⎡ 1 0 0 ⎦ ⎤
for λ 2 = 2 , i s X 2 = k [ 0 1 1 ] \lambda_{2}=2,\hspace{0.2cm}is\hspace{0.2cm}X_{2}=k\begin{bmatrix}
0 \\
1\\
1
\end{bmatrix} λ 2 = 2 , i s X 2 = k ⎣ ⎡ 0 1 1 ⎦ ⎤
for λ 3 = 1 , i s X 3 = k [ 0 1 − 1 ] \lambda_{3}=1,\hspace{0.2cm}is\hspace{0.2cm}X_{3}=k\begin{bmatrix}
0 \\
1\\
-1
\end{bmatrix} λ 3 = 1 , i s X 3 = k ⎣ ⎡ 0 1 − 1 ⎦ ⎤
Clearly, the eigen vectors X 1 = [ 1 0 0 ] X_{1}=\begin{bmatrix}
1 \\
0\\
0
\end{bmatrix} X 1 = ⎣ ⎡ 1 0 0 ⎦ ⎤ , X 2 = [ 0 1 1 ] X_{2}=\begin{bmatrix}
0 \\
1\\
1
\end{bmatrix} X 2 = ⎣ ⎡ 0 1 1 ⎦ ⎤ and X 3 = [ 0 1 − 1 ] X_{3}=\begin{bmatrix}
0 \\
1\\
-1
\end{bmatrix} X 3 = ⎣ ⎡ 0 1 − 1 ⎦ ⎤ are linearly independent and orthogonal.
The normalized vectors of the above eigen vextors X 1 , X 2 , X 3 X_{1}, X_{2},X_{3} X 1 , X 2 , X 3 are:
e 1 = X 1 ∣ ∣ X 1 ∣ ∣ = [ 1 0 0 ] e_{1}=\frac{X_{1}}{\mid\mid X_{1}\mid\mid}= \begin{bmatrix}
1 \\
0\\
0
\end{bmatrix} e 1 = ∣∣ X 1 ∣∣ X 1 = ⎣ ⎡ 1 0 0 ⎦ ⎤ , e 2 = X 2 ∣ ∣ X 2 ∣ ∣ = [ 0 1 2 1 2 ] \displaystyle e_{2}=\frac{X_{2}}{\mid\mid X_{2}\mid\mid}= \begin{bmatrix}
0 \\
\frac{1}{\sqrt2}\\
\frac{1}{\sqrt2}
\end{bmatrix} e 2 = ∣∣ X 2 ∣∣ X 2 = ⎣ ⎡ 0 2 1 2 1 ⎦ ⎤ , e 3 = X 3 ∣ ∣ X 3 ∣ ∣ = [ 0 1 2 − 1 2 ] \displaystyle e_{3}=\frac{X_{3}}{\mid\mid X_{3}\mid\mid}= \begin{bmatrix}
0 \\
\frac{1}{\sqrt2}\\
-\frac{1}{\sqrt2}
\end{bmatrix} e 3 = ∣∣ X 3 ∣∣ X 3 = ⎣ ⎡ 0 2 1 − 2 1 ⎦ ⎤
Therefore,
The normalized model matrix is
p ˉ = [ 1 0 0 0 1 2 1 2 0 1 2 − 1 2 ] \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} p ˉ = ⎣ ⎡ 1 0 0 0 2 1 2 1 0 2 1 − 2 1 ⎦ ⎤ and p ˉ − 1 = p ˉ T = 1 2 [ 2 0 0 0 1 1 0 1 − 1 ] \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} p ˉ − 1 = p ˉ T = 2 1 ⎣ ⎡ 2 0 0 0 1 1 0 1 − 1 ⎦ ⎤
p ˉ − 1 A p ˉ = 1 2 [ 2 0 0 0 1 1 0 1 − 1 ] [ 1 0 0 0 3 − 1 0 − 1 3 ] [ 1 0 0 0 1 2 1 2 0 1 2 − 1 2 ] \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} p ˉ − 1 A p ˉ = 2 1 ⎣ ⎡ 2 0 0 0 1 1 0 1 − 1 ⎦ ⎤ ⎣ ⎡ 1 0 0 0 3 − 1 0 − 1 3 ⎦ ⎤ ⎣ ⎡ 1 0 0 0 2 1 2 1 0 2 1 − 2 1 ⎦ ⎤
= [ 1 0 0 0 1 1 0 1 − 1 ] = D =\begin{bmatrix}
\ 1 & 0 & 0 \\
0 & 1 & 1 \\
0 & 1 & -1
\end{bmatrix}=D = ⎣ ⎡ 1 0 0 0 1 1 0 1 − 1 ⎦ ⎤ = D
We know that the quadratic form is Q = X T A X Q=X^{T}AX Q = X T A X where X = P ˉ Y X=\bar{P}Y X = P ˉ Y
then,
Q = ( P ˉ Y ) A ( P ˉ Y ) Q=(\bar{P}Y)A(\bar{P}Y) Q = ( P ˉ Y ) A ( P ˉ Y )
= ( Y T P ˉ T ) A ( P ˉ Y ) = Y T ( P T A P ˉ ) Y = Y T ( P ˉ − 1 A P ˉ ) Y s i n c e P ˉ T = P ˉ − 1 = Y T D Y =(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 = ( Y T P ˉ T ) A ( P ˉ Y ) = Y T ( P T A P ˉ ) Y = Y T ( P ˉ − 1 A P ˉ ) Y s in ce P ˉ T = P ˉ − 1 = Y T D Y
Therefore, the canonical form
= [ y 1 y 2 y 3 ] [ 1 0 0 0 2 0 0 0 4 ] [ y 1 y 2 y 3 ] =\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} = [ y 1 y 2 y 3 ] ⎣ ⎡ 1 0 0 0 2 0 0 0 4 ⎦ ⎤ ⎣ ⎡ y 1 y 2 y 3 ⎦ ⎤
= y 1 2 + 2 y 2 2 + 4 y 3 2 = y_{1}^{2}+2y_{2}^{2}+4y_{3}^{2} = y 1 2 + 2 y 2 2 + 4 y 3 2
which gives the canonical form.
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