Answer to Question #288887 in Linear Algebra for Nathiya

Question

2x square + y square+z square +4 yz +2 xy-2 xx

Answer 1

$2x^{2}+y^{2}+z^{2}+4yz+2xy-2xz$

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

$\displaystyle Q(x)=2x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+4x_{2}x_{3}+2x_{1}x_{2}-2x_{1}x_{3}$

The symmetric matrix A with the given quadratic form is

$A=$ $\begin{bmatrix} 2 & 1 & -1 \\ 1 & 1 & 2\\ -1 & 2 & 1 \end{bmatrix}$

The characteristic equation of A is

$\mid(A-\lambda I)\mid=-\lambda^{3}+4\lambda^{2}+\lambda-12=0\\\implies \lambda=3,\frac{-1+17}{2},\frac{1+17}{2}$

The corresponding eigen values are:

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

for $\lambda_{2}=\frac{-1+17}{2},\hspace{0.2cm}is\hspace{0.2cm}X_{2}=\begin{bmatrix} \frac{-3+17}{2} \\ -1\\ 1 \end{bmatrix}$

for $\lambda_{3}=1,\hspace{0.2cm}is\hspace{0.2cm}X_{3}=\begin{bmatrix} -\frac{3+17}{2} \\ -1\\ 1 \end{bmatrix}$

Clearly, the eigen vectors $X_{1}=\begin{bmatrix} 0 \\ 1\\ 1 \end{bmatrix}$ , $X_{2}=\begin{bmatrix} \frac{-3+17}{2} \\ -1\\ 1 \end{bmatrix}$ and $X_{3}=\begin{bmatrix} -\frac{3+17}{2} \\ -1\\ 1 \end{bmatrix}$ are linearly independent and orthogonal.