Answer to Question #288172 in Calculus for Sajjala Hemanth ku

Question

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

Answer 1

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

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

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

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

The symmetric matrix A with the given quadratic form is

$A=$ $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 3 & -1\\ 0 & -1 & 3 \end{bmatrix}$

The characteristic equation of A is

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

The corresponding eigen values are:

for $\lambda_{1}=1,\hspace{0.2cm}is\hspace{0.2cm}X_{1}=k\begin{bmatrix} 1 \\ 0\\ 0 \end{bmatrix}$

for $\lambda_{2}=2,\hspace{0.2cm}is\hspace{0.2cm}X_{2}=k\begin{bmatrix} 0 \\ 1\\ 1 \end{bmatrix}$

for $\lambda_{3}=1,\hspace{0.2cm}is\hspace{0.2cm}X_{3}=k\begin{bmatrix} 0 \\ 1\\ -1 \end{bmatrix}$

Clearly, the eigen vectors $X_{1}=\begin{bmatrix} 1 \\ 0\\ 0 \end{bmatrix}$ , $X_{2}=\begin{bmatrix} 0 \\ 1\\ 1 \end{bmatrix}$ and $X_{3}=\begin{bmatrix} 0 \\ 1\\ -1 \end{bmatrix}$ are linearly independent and orthogonal.

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

$e_{1}=\frac{X_{1}}{\mid\mid X_{1}\mid\mid}= \begin{bmatrix} 1 \\ 0\\ 0 \end{bmatrix}$ , $\displaystyle e_{2}=\frac{X_{2}}{\mid\mid X_{2}\mid\mid}= \begin{bmatrix} 0 \\ \frac{1}{\sqrt2}\\ \frac{1}{\sqrt2} \end{bmatrix}$ , $\displaystyle e_{3}=\frac{X_{3}}{\mid\mid X_{3}\mid\mid}= \begin{bmatrix} 0 \\ \frac{1}{\sqrt2}\\ -\frac{1}{\sqrt2} \end{bmatrix}$

Therefore,

The normalized model matrix is

$\displaystyle \bar{p}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}} \\ 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}$ and $\bar{p}^{-1}=\bar{p}^{T}=\frac{1}{\sqrt{2}}\displaystyle \begin{bmatrix} \sqrt{2} & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & - 1 \end{bmatrix}$

$\bar{p}^{-1}A\bar{p}= \frac{1}{\sqrt{2}}\displaystyle \begin{bmatrix} \sqrt{2} & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & - 1 \end{bmatrix}\begin{bmatrix} \ 1 & 0 & 0 \\ 0 & 3 & -1 \\ 0 & -1 & 3 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}} \\ 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}$

$=\begin{bmatrix} \ 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & -1 \end{bmatrix}=D$

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

then,

$Q=(\bar{P}Y)A(\bar{P}Y)$

$=(Y^{T}\bar{P}^{T})A(\bar{P}Y)\\=Y^{T}(P^{T}A\bar{P})Y\\=Y^{T}(\bar{P}^{-1}A\bar{P})Y\hspace{0.3cmsince\hspace{0.3cm}}\bar{P}^{T}=\bar{P}^{-1}\\=Y^{T}DY$

Therefore, the canonical form

$=\begin{bmatrix} y_{1} & y_{2} &y_{3} \\ \end{bmatrix} \begin{bmatrix} \ 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{bmatrix} \begin{bmatrix} \ y_{1} \\ y_{2} \\ y_{3} \end{bmatrix}$

$= y_{1}^{2}+2y_{2}^{2}+4y_{3}^{2}$

which gives the canonical form.