Canonic form of a quadric

Question

I have a question, how can determine the canonic form of the quadric associated to $F$, where $F$ is a bilinear form $\Bbb R^3\times\Bbb R^3\to\Bbb R $ that its associated matrice is $A=\left(\begin{matrix}0&0&-1\\ 0&-1&0\\ -1&0&0\\ \end{matrix}\right)$?

Another question, what is the base of $\Bbb R^3$ in whic the quadric is in canonic form?

Answer 1

The quadratic form associated to $F$ is $x^TAx$ and since $A$ is symmetric by spectral theorem $A$ can be diagonalized by a basis of orthogonal eigenvectors $v_1,v_2,v_3$ that is

$$M=[v_1\,v_2\,v_3] \implies x=My\quad M^{-1}=M^T$$

and then

$x^TAx=(My)^TA(My)=y^TM^TAMy=y^T(M^{-1}AM)y=y^TDy=\lambda_1y_1^2+\lambda_2y_2^2+\lambda_3y_3^2$

which is the canonical form.

Answer 2

Find an orthonormal basis $(v_1,v_2,v_3)$ of $\mathbb{R}^3$ such that each $v_k$ is an eigenvector of $A$. If the correcponding eigenvalues are $\lambda_1$, $\lambda_2$, and $\lambda_3$, then the canonical form will be $$(X_1,X_2,X_3)\mapsto\lambda_1{X_1}^2+\lambda_2{X_2}^2+\lambda_3{X_3}^2.$$