Answer to Question #294190 in Linear Algebra for Navab

Question

Reduce the quadratic form 2𝑥

2 + 2𝑦

2 + 2𝑧

2 + 2𝑦𝑧 to the canonical form by 

orthogonal reduction. Find the index, signature and nature of the quadratic form.

Answer 1

Matrix A$=\begin{pmatrix} 2&0&0 \\ 0&2&1 \\ 0&1&2 \end{pmatrix}$

$\>\>\begin{vmatrix} A-\lambda\>I \\ \end{vmatrix}=\begin{vmatrix} 2-\lambda&0&0 \\ 0&2-\lambda& 1\\ 0&1&2-\lambda \end{vmatrix}=0$

$(2-\lambda)\begin{bmatrix} (2-\lambda)(2-\lambda)-1\\ \end{bmatrix}+0+0=0$

$(2-\lambda)(\lambda^2-4\lambda+3)=0$

$(2-\lambda)(\lambda-1)(\lambda-3)=0$

$\lambda=1,\lambda=2,\lambda=3$

For $\lambda=1,\begin{pmatrix} 1&0&0 \\ 0&1&1 \\ 0&1&1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix} 0 \\ 0\\ 0 \end{pmatrix}$

$x_1=0,\>x_2=-1,\>x_3=1$

For $\lambda=2\begin{pmatrix} 0&0 & 0 \\ 0&0 & 1\\ 0&1&0 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$

$x_1=1,\>x_2=0,\>x_3=0$

For $\lambda=3\begin{pmatrix} -1&0&0 \\ 0&-1 & 1\\ 0&1&-1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2\\ x_3 \end{pmatrix}=\begin{pmatrix} 0 \\ 0\\ 0 \end{pmatrix}$

$x_1=0,\>x_2=1,\>x_3=1$

Modal matrix $\begin{pmatrix} 0&1&0 \\ -1&0&1\\ 1&0&1 \end{pmatrix}$

Normalized modal matrix

$Q=\begin{pmatrix} 0&1&0 \\ \frac{-1}{\sqrt2} & 0&\frac{1}{\sqrt2}\\ \frac{1}{\sqrt2}&0&\frac{1}{\sqrt2} \end{pmatrix}$

Diagonalizing matrix

$D=Q^TAQ=\begin{pmatrix} 0&\frac{-1}{\sqrt2}& \frac{1}{\sqrt2} \\ 1&0&0 \\ 0&\frac{1}{\sqrt2}&\frac{1}{\sqrt2} \end{pmatrix}\begin{pmatrix} 2&0&0 \\ 0&2 & 1\\ 0&1&2 \end{pmatrix}\begin{pmatrix} 0&1&0 \\ \frac{-1}{\sqrt2}&0&\frac{1}{\sqrt2} \\ \frac{1}{\sqrt2}&0&\frac{1}{\sqrt2} \end{pmatrix}$

$=\begin{pmatrix} 1&0&0 \\ 0&2& 0\\ 0&0&3 \end{pmatrix}$

Diagonalized matrix has principal diagonal element =Eigenvalues

All other elements$=0$

The orthogonal transformation reduces the quadratic form to conical form

$y_1^2+2y_2^2+3y_3^2$

Index$=3$

Signature$=2×3-3=3$

Nature of quadratic form

Positive definate