Let $q_1(x)$ and $q_2(x)$, be two quadratic forms on a vector space $V$ based on $\mathbb{R}$. In the base $\mathcal{B} = (e_1, e_2, e_3)$, the expressions of the two forms are:
$$ \begin{align} & q_1(x) = 3x_1^2 - 24x_1x_2 + 12x_1x_3 + 48x_2^2 - 52x_2x_3 + 12x_3^2 \\ & q_2(x) = x_1^2 + 2x_1x_2 + 2x_2^2 - 2x_2x_3 - x_3^2 \end{align} $$
I have to give the expression of $\lambda : V \to V : x \mapsto \lambda(x)$, an isometry of $(V,q_1)$ in $(V,q_2)$.
First, I figured out that $q_1$ and $q_2$ was the same quadratic form $q$, expressed in two different basis, because of the signature of $q_1$ and $q_2$. Thus, find an isometry is equivalent to find $\lambda$ represented in $\mathbb R^3_3$ by the matrix $\Lambda$, such that this equality holds true:
$$ G_1 = \Lambda^TG_2\Lambda $$
Where $G_i$ is the matrix representing $g_{q_i}$, the polar form of $q_i$, in the base $\mathcal B$.
1) Is my reasoning correct about the the quadratic form $q$ and the isometry $\lambda$ ?
2) If it is true, how can I find the matrix $\Lambda$ ?
3) If it is false, how am I supposed to give the expression of $\lambda$ ?
Thank you very much for your help!