# Properties of non-zero symmetric bilinear form

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

1. rank $f=1$
2. $\dim\{\beta\in\Bbb R^3:f(\alpha,\beta)=0\text{ for all }\alpha\in\Bbb R^3\}=2$
3. $f$ is a positive semi-definite or negative semi-definite
4. $\{\alpha:f(\alpha,\alpha)=0\}$ 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?

• The maps cannot have trivial kernel: By the Rank Nullity Thorem, $\dim \operatorname{im} T_i + \dim \ker T_i = 3$, but $\dim \operatorname{im} T_i \leq \dim \Bbb R = 1$, so $\dim \ker T_i \geq 2$. – Travis Nov 25 '15 at 13:06
• @Travis, and since $f$ is non-zero map, $\dim \ker T_i\ne 3$, and hence $\dim \ker T_i=2$ and $\dim \ker f=2$, showing (1),(2) and (4) are true! Am I in the right path? What about (3)? – Jesse P Francis Nov 25 '15 at 17:53
• There are some intermediate arguments to make for (2) and (4), but yes, that's right idea. In fact, (3) is automatically true for any rank $1$ bilinear form: In some basis, the form has matrix representation of the form $\operatorname{diag}(\lambda, 0, \ldots, 0)$. – Travis Nov 25 '15 at 22:19
• @Travis, I have posted an answer, can you please verify if my arguments are correct? – Jesse P Francis Jun 15 '16 at 6:50

## 1 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 $\text{diag}(\lambda,0,…,0)$