How can we find the space $x : x^\top T x = 0$ Given symmetric matrix $T$ with real entries, is there a way to find the space of all $x$ such that $$x^\top T x =0$$?
This is superficially similar to the null space of $T$, so I hope methods similar to those for finding the null space may exist.  I have explored that avenue without good results so far.  My only thought, unresolved so far, is that this is the space of vectors related to the zero eigenvalue.
Is there a name for this property that $x$ must have relative to $T$, for $x^\top T x=0$?
 A: Denote $Z:=\{x: \ x^TTx=0\}$.
This set is not equal to the set of eigenvectors to the zero eigenvalue.
Consider $T=\pmatrix{1&0\\0&-1}$, then
$$
x^TTx=0
$$
if and only if $|x_1|=|x_2|$. This is fulfilled for, e.g., $x=\pmatrix{1\\1}$, which is not an eigenvector of $T$.
The set of such vectors $x$ is also not a vector space. One can only prove
that $Z$ is closed with respect to scalar multiplication
$$
x\in Z\Rightarrow \lambda x\in Z\ \forall \lambda.
$$
Closedness with respect to addition of vectors does not hold in general.
The set $Z$ is trivial ($=\{0\}$) if and only if $T$ is positive definite or negative definite. If $T$ has positive and negative eigenvalues, then $Z$ is non-trivial ($\ne\{0\}$)by the mean-value theorem.
A: For matrix $T$, we have eigenvectors $v_i$ and corresponding eigenvalues $\lambda_i$.
Theorem: given vector $\bf a$ such that $$\sum_i a_i^2 \lambda_i = 0,$$ all $x:=\sum_i a_i v_i$ satisify $x^\top T x = 0$.
Proof: Since $v_i$ are eigenvectors of $T$, and since $T$ is symmetric with real entries the eigenvectors are orthonormal,
$$ x^\top T x 
= \left(\sum_i a_i v_i \right)^\top T \left(\sum_i a_i v_i\right) 
= \left(\sum_i a_i v_i \right)^\top \left(\sum_i a_i v_i \lambda_i \right)
= \sum_i a_i^2 \langle v_i, v_i \rangle \lambda_i
= \sum_i a_i^2 \lambda_i = 0.
$$
Q.E.D.
Since $\bf a$ can be constructed from the null space of $\bf \lambda$ by restricting it to the quadrant of nonnegative values, we can also construct $x$ this way.
