How can i find all isotropic vectors of quadratic form
$x_{1}^2+4x_{2}^2+8x_{3}^2-4x_{1}x_{2}+8x_{1}x_{3}-14x_{2}x_{3}$
I don't understand how to approach.
Isotropic vectors of quadratic form
linear algebraquadratic-forms
Related Solutions
Let $V$ be a maximal totally isotropic subspace. Let $W$ be a subspace of dimension $\max\{r,s\}$ on which $Q$ is positive or negative definite. (For instance, if $\max\{r,s\}=r$, let $W$ be the span of $e_1,\ldots,e_r$, and if $\max\{r,s\}=s$, let $W$ be the span of $e_{r+1},\ldots, e_s$.) Then \begin{equation*} \dim(V\cap W) = \dim(V) + \dim(W) - \dim(V+W) \end{equation*} is a basic result from linear algebra. Since $V$ is totally isotropic, $V\cap W=0$, so we get \begin{align*} \dim(V) &= \dim(V+W)-\dim(W) \\ &\leq \dim(X)-\max\{r,s\} \\ &= r + s - \max\{r,s\} \\ &= \min\{r,s\}. \end{align*} On the other hand, certainly we have $\dim(V)\geq\min\{r,s\}$, because $V$ is a maximal totally isotropic subspace and we know that there exists a totally isotropic subspace with dimension $\min\{r,s\}$.
Therefore $\dim(V)=\min\{r,s\}$.
EDIT: @JanVysoky is correct, I was too glib in claiming the inequality $\dim(V)\geq\min\{r,s\}$. Let $B$ be the symmetric bilinear form on $X$ corresponding to the quadratic form $Q$. Consider the subspace $V^\perp$ defined by \begin{equation*} V^\perp = \{v^\perp\in X:B(v,v^\perp)=0\text{ for all } v\in V\}. \end{equation*} The theory of bilinear forms tells us that, since $Q$ and hence $B$ is non-degenerate, $\dim(V^\perp) = \dim(X)-\dim(V)$. Let $W$ be a totally isotropic subspace of dimension $\min\{r,s\}$. (For instance, if $\min\{r,s\}=s$, let $W$ be the span of $e_1+e_{r+s},e_2+e_{r+s-1},\ldots,e_s+e_{r+1}$ as @Bach said in the comments.) The maximality of $V$ implies $V^\perp\cap W=0$. Then \begin{align*} 0 &= \dim(V^\perp\cap W) \\ &= \dim(V^\perp) + \dim(W) - \dim(V^\perp + W) \\ &\geq (\dim(X) - \dim(V)) + \dim(W) - \dim(X) \\ &= \min\{r,s\} - \dim(V) \end{align*} Therefore $\dim(V)\geq\min\{r,s\}$.
there is a simple procedure for finding all rational vectors in the null cone, given $q(z) = 0.$ Let $$ 2 q(x) = x^T H x, $$ where $H$ is the Hessian matrix, all integers. Next, take any primitive integer vector $v$ (we need to assume that $q(v) \neq 0$ and find the value of real $t$ such that $q(z + tv) = 0.$ It turns out that $$ t = \frac{-2z^T H v}{v^T H v} $$ which is the reason that we needed $q(v) \neq 0.$ Now we have a rational null vector $$ \frac{(v^THv)z -2(z^T Hv)v}{v^T Hv} $$
To get an integer null vector we may now just multiply by the denominator $q(v).$ The trouble is that the result will often be an integer vector that is not primitive, the GCD of the entries is something larger than one. Well, find the gcd and divide out by it.
So, take column vector $v,$ if $q(v) \neq 0$ find $$ (v^THv)z -2(z^T Hv)v $$ and divide through by its gcd of entries.
In low dimension there are ways to get around the gcd business. the familiar example would be Pythagorean triples, where primitive null vectors are found parametrized by three binary quadratic forms. I can be specific in dimension 3: given $H$ the Hessian of an integer isotropic form, and the matrix $W$ as the Hessian of $y^2 - zx,$ there is an invertible integer matrix $P$ such that $P^T HP = n W.$ Then recipes can be given for all null vectors.
Fricke and Klein (1897), in dimension 4, use $x^2 + y^2 + z^2 - w^2$ for signature $+++-,$ then $xy - zw$ for signature $++--.$
Best Answer
Use Lagrange's Reduction (complete the square). I'll use $\;x,y,z\;$ for ease:
$$x^2+4y^2+8z^2-4xy+8xz-14yz=(x^2-2x(2y+4z))+4y^2+8z^2-14yz=$$
$$=\left(x-2y-4z\right)^2-\color{red}{4y^2}-16yz-16z^2+\color{red}{4y^2}+8z^2-14yz=$$
$$=\left(x-2y-4z\right)^2-8\left(z^2+\frac{15}2yz\right)=(x-2y-4z)^2-8\left(z+\frac{15}4y\right)^2+\frac{225}2y^2$$
Thus, your quadratic is equivalent to $\;\tilde x^2+\tilde y^2-\tilde z^2\;$ (signature $\,(2,1)\,$) , so it actually is a cone.
$$$$