Let $f$ be a non-zero symmetric bilinear form on $\Bbb R^3$. Suppose that there exist linear transformations $T_i:\Bbb R^3\to\Bbb R, i=1,2$ such that for all $\alpha,\beta\in\Bbb R^3,f(\alpha,\beta)=T_1(\alpha)T_2(\beta)$. Then
- rank $f=1$
- $\dim\{\beta\in\Bbb R^3:f(\alpha,\beta)=0\text{ for all }\alpha\in\Bbb R^3\}=2$
- $f$ is a positive semi-definite or negative semi-definite
- $\{\alpha:f(\alpha,\alpha)=0\}$ is a linear subspace of dimension $2$
(CSIR December 2014)
My attempt:
(1), (2) and (4) are wrong if $\ker(T_1)=\{0\},\ker(T_2)=\{0\}$
All information I could find about positive/negative semi-definiteness of bilinear form is this post about positive definiteness.
Edit: (question has already been answered in comments; "intermediate link" Travis talks about is here.
Can someone please guide me?
Best Answer
Answering my own question with hints from Travis: $$\dim \text{im }{T_i}\le1$$
Since $f$ is non zero, we have $$\dim \text{im }{T_i}=1$$
Hence $$\dim \text{Ker }T_i=3-1=2$$
Hence, $$\text{rank}(f)=1$$
$$\begin{array}1\dim\{\beta\in\Bbb R^3:f(\alpha,\beta)=0 \forall\alpha\in\Bbb R^3\}&&=\dim\{\beta\in\Bbb R^3:T_1(\alpha)T_2(\beta)=0 \forall\alpha\in\Bbb R^3\}\\ &&=\dim\{\beta\in\Bbb R^3:T_2(\beta)=0\}\\&&=2\end{array}$$
and similarly, $$\begin{array}1\dim\{f(\alpha,\alpha)=0\}&&=\dim\{T_1(\alpha)\text{ or }T_2(\alpha)=0\}\ge2\end{array}$$ It cannot be equal to $3$ since in that case $f$ will be a zero map. Therefore, we have
$$\begin{array}1\dim\{f(\alpha,\alpha)=0\}=2\end{array}$$
(3) is automatically true for any rank 1 bilinear form: In some basis, the form has matrix representation of the form $\text{diag}(\lambda,0,…,0)$