Answer to Question #171369 in Linear Algebra for Kings

Question

reduce the following quadratic form 3x²-2y²-z²+12yz-4xy+8zx to canonical form by an orthogonal transformation. Also find the rank, index signature and nature

Answer 1

CORRECTED SOLUTION

The quadratic form 3x²-2y²-z²+12yz-4xy+8zx has the matrix

$A=\begin{pmatrix} 3 & -2 & 4\\ -2 & -2 & 6\\ 4 &6 & -1 \end{pmatrix}$

Its characteristic polynomial is

$\det \begin{pmatrix} 3-\lambda & -2 & 4\\ -2 & -2-\lambda & 6\\ 4 &6 & -1-\lambda \end{pmatrix}=-\lambda^3 +63\lambda-162$

One can notice that $\lambda=3$ is a root of this polynomial.

$\frac{\lambda^3-63\lambda+162}{\lambda-3}=\lambda^2+3\lambda-54=(\lambda+9)(\lambda-6)$

Therefore, $\lambda_1=3$, $\lambda_2=6$, $\lambda_3=-9$ are all the eigenvalues of the given matrix.

Find the corresponding eigenvectors by solving the equations $AX=\lambda_iX$ with i=1,2,3.

1) Case $\lambda_1=3$. The equation $AX=3X$ implies

$-2y+4z=0$

$-2x-5y+6z=0$

The fundamental solution of this system is $z=t, y=2t, x=-2t$.

Therefore, the first eigenvector is $e_1=(-2,2,1)^T$. $|e_1|=\sqrt{(-2)^2+2^2+1^2}=3$

2) Case $\lambda_2=6$. The equation $AX=6X$ implies

$-3x-2y+4z=0$

$-2x-8y+6z=0$

The fundamental solution of this system is $z=\begin{vmatrix} -3 & -2 \\ -2 & -8 \end{vmatrix}t=20t, y=\begin{vmatrix} 4 & -3 \\ 6 & -2 \end{vmatrix}t=10t, x=\begin{vmatrix} -2 & 4 \\ -8 & 6 \end{vmatrix}t=-20t$.

Therefore, the second eigenvector is $e_2=(-2,1,2)^T$. $|e_2|=\sqrt{(-2)^2+1^2+2^2}=3$

3) Case $\lambda_3=-9$. The equation $AX=-9X$ implies

$12x-2y+4z=0$

$-2x+7y+6z=0$

The fundamental solution of this system is

$z=\begin{vmatrix} 12 & -2 \\ -2 & 7 \end{vmatrix}t=-80t, y=\begin{vmatrix} 4 & 12 \\ 6 & -2 \end{vmatrix}t=-80t, x=\begin{vmatrix} -2 & 4 \\ 7 & 6 \end{vmatrix}t=-40t$.

Therefore, the third eigenvector is $e_3=(1,2,2)^T$. $|e_3|=\sqrt{1^2+2^2+2^2}=3$

In the orthonormal basis $e_1/3, e_2/3, e_3/3$ the quadratic form will be

$3y_1^2+6y_2^2-9y_3^2$

This is the normal form of the given quadratic form. From it one can see that the quadratic form is non-degenerate, indefinite, its rank=3, its index (the number of positive terms) =2 and its signature (the difference between the number of positive terms and the number of negative terms) is 1.