find real numbers so that a specific linear isometry exists Consider the quadratic form
$$f: \mathbb{R}^3 \to \mathbb{R}, (x_1, x_2, x_3) \mapsto 3x_1^2 - 3x_2^2 + x_3^2-2x_1x_3$$
I want to find $\lambda_1, \lambda_2, \lambda_3 \in \mathbb{R}$, so that there exists a linear isometry $\alpha$ of the Euclidian space $\mathbb{R}^3$, so that $f \circ \alpha(x_1, x_2, x_3) = \lambda_1x_1^2 + \lambda_2x_2^2 + \lambda_3x_3^2$. Also, how does the set $A := \{x \in \mathbb{R}^3: f \circ \alpha(x) = 0\}$ look like?
I thought that the first step might be finding an orthonormal basis for $(\mathbb{R}^3, f)$, and then somehow playing around with the change of basis matrices, but I don't really know what would come after that.
 A: The matrix of $f$ w.r.t. the base whose coordinates are $x_1,x_2,x_3$ is $$ F = \left( \begin{array}{ccc}
3 & 0 & -1 \\
0 & -3 & 0  \\
-1 & 0 & 1 \end{array} \right) \, .$$
That is to say $$ f(x_1,x_2,x_3) = (x_1 \, x_2 \, x_3) \left( \begin{array}{ccc}
3 & 0 & -1 \\
0 & -3 & 0  \\
-1 & 0 & 1 \end{array} \right)  \left( \begin{array}{c}
x_1 \\
x_2 \\
x_3 \end{array} \right) \, .$$
The above is a symmetric $3 \times 3$ matrix with eigenvalues $-3, 2 \pm \sqrt{2}$. Let $P$ be the matrix whose first column is a unit eigenvector relative to $-3$, the second column is a unit eigenvector relative to $2 - \sqrt{2}$ and the third column a unit eigenvector of $2 + \sqrt{2}$. The $P$ is an orthogonal matrix i.e. $P^{\top} = P^{-1}$ and $$P^{\top} F P = \left( \begin{array}{ccc}
-3 & 0 &  \\
0 & 2 - \sqrt{2} & 0  \\
0 & 0 & 2 + \sqrt{2} \end{array} \right) $$
This means that $P$ (resp. $-3, 2-\sqrt{2}, 2 + \sqrt{2}$) is the $\alpha$  (resp. $\lambda_1,\lambda_2,\lambda_3$) you are looking for.
Since one of the $\lambda$'s is negative and the other two positive your set $A$ looks like the light cone https://en.wikipedia.org/wiki/Light_cone
A: Use this Gauss decomposition in squares: $f(y_1,y_2,y_3)=3(y_1-1/3y_3)^2-3y_2^2+2/3y_3^2$.
