Let f be non-zero symmetric bilinear form on $\mathbb{R}^3.$ Suppose there exist linear transformation$ T_i:\mathbb{R}^3\to\mathbb{R}, i=1,2 $ such that for all $a,b\in \mathbb{R}^3:f(a,b)=T_1(a)T_2(b)$ .
Then:
1.$\ rank\ f=1$
2.$\ \dim \{b\in \mathbb{R}^3| \forall{a \in \mathbb{R}^3}: f(a,b)=0 \}$
3.$\ f$ is a positive semi-definite or negative semi-definite.
4.$\ \{a\in \mathbb{R}^3:f(a,a)=0 \}$ is a linear subspace of dimension 2.
rank f is 1 by observing matrix of the transformation but I don't know how to calculate dimension of $\{b\in \mathbb{R}^3|\forall{a \in \mathbb{R}^3}:f(a,b)=0\}$. And what about last option. I guess third option is correct. As matrix of f has rank 1. Is my logic is correct? Please guide me.
Preliminaries:
I treat the case $T_1=T_2=T$. You will have no difficulty to apply it to the general case.
Linear form $T$ has a $1 \times 3$ matrix, $(u,v,w)$.
Let us denote by $V$ the column vector $V=\begin{pmatrix}u\\v\\w\end{pmatrix}$.
We then have:
$$T_1(A)=V^TA.$$
Thus
$$T_1(A)T_1(B)=(A^TV)(V^TB)=A(VV^T)B$$
Therefore the quadratic form has this matrix:
$$Q=VV^T=\begin{pmatrix}u\\v\\w\end{pmatrix} \begin{pmatrix}u&v&w\end{pmatrix}=\left(\begin{array}{rrr}u^2 & uv & uw\\ uv & v^2 & vw\\ uw & vw & w^2\end{array}\right) $$
Question 1: $Q$ is a rank one matrix because its image space (column space) is directed by the unique vector $V$ (first column is $uV$, second column is $vV$, third column is $wV$).
Question 2: if you write the equations, you will find a 2 dimensional solution.
Question 3: $$X^TQX=X^TVV^TX=(V^TX)^T(V^TX)=\|V^TX\|^2 \geq 0$$
with the possibility that the result is zero when $X$ is orthogonal to $V$.
Question 4: Write down the equations resulting from constraint $f(A,A)=0$, i.e., $A^TQA=0$, i.e.,
$$\begin{pmatrix}a&b&c\end{pmatrix}\begin{pmatrix}u\\v\\w\end{pmatrix} \begin{pmatrix}u&v&w\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=0 \iff$$
$$(au+bv+cw)^2=0 \ \iff \ au+bv+cw=0,$$
clearly a dimension 2 subspace called the isotropic space associated with the quadratic form $Q$.