When is $f(a, b) = h(a)h(b)$ a non-singular bilinear map, where $h$ is linear? 
Let $M$ be a vector space over the field $\Bbb{R}$ and $h$ : $M \to \Bbb{R}$ be a linear map.
Define $f : M \times M  \to \Bbb{R}$ such that $f(x, y) = h(x)h(y)$. Then show that

*

*$f$ is a bilinear map (I did it)

*Under what conditions $f$ is non-singular bilinear map?


I need a help in the second part. I did the following:
To show that $f$ is non-singular, which means ${M}^\bot = \{0\}$, where
$$M^\bot = \{ b \in M \mid \forall a \in M, f(a, b) = \{0\}\},$$
$f(a, b) = h(a)h(b) = \{0\}$ and since $h(a), h(b) \in \Bbb{R}$
then $h(a)h(b) = \{0\}$ iff $h(a) = \{0\}$ or $h(b) = \{0\}$.
Case 1: If $h(b) = \{0\} = h(\{0\})$ and if $h$ is 1-1 map then $b = 0$.
Case 2: if $h(a) = \{0\}$ and we assume that $h(x) = \{0\}$ iff $x = 0$ then $a = 0$.
So $M^\bot = \{0\}$ iff $h$ is 1-1 map and $h(x) = \{0\}$ iff $x = 0$.
Is it true?
 A: If $\dim(M)=0$, then $M=\{0\}$ so $M^\bot=\{0\}$ is trivially true in this case and $f$ is non-degenerate. Also $h:M\to\mathbb R$ is obviously 1-1. From now on, we assume $\dim(M)>0$.

Method 1 (idea of the OP)
We want to show that

$f$ is non-degenerate if and only if $\dim(M)=1$ and $h$ 1-1.

We prove the implication "if $f$ is non-degenerate, then $\dim(M)=1$ and $h$ is 1-1" by contraposition. Assume that $\dim(M)>1$ or $h$ is not 1-1. Note that, if $\dim(M)>1$, then $h$ is not 1-1. This means that there is $b\neq 0$ such that $h(b)=0$. Then, for every $a$ in $M$, we have $f(a,b)=h(a)h(b)= 0$ so $b\in M^\bot$.  This means that $f$ is degenerate. So the contraposition is true.
Conversely, assume that $h$ is 1-1 and $\dim(M)=1$. Then $h(a)=ma$ for some $m\in\mathbb R\setminus\{0\}$ and we have $f(a,b)=m^2ba$. Clearly, $M^\bot=\{0\}$ (if $b\neq 0$, then $f(1,b)=m^2b\neq 0$) so $f$ is non-degenerate.

Alternative method (in the case where $\dim(M)$ is finite):
Assume that $n=\dim(M)>0$ and let's find the matrix of the bilinear form $f(a,b)=h(a)h(b)$. Fix a basis $\mathcal B$ of $M$. $h$ is a linear form from $M$ into $\mathbb R$ so there exists a $n\times 1$ matrix $H$ such that $h(a)=Ha$ (where $a$ is represented as a column vector in the basis $\mathcal B$). We now have
$$
f(a,b) = h(a)h(b) = HaHb = (Ha)^T Hb = a^T (H^TH) b
$$
so the matrix of the bilinear form $f$ in the basis $\mathcal B$ is $S=H^TH$ (note that $(Ha)^T=Ha$ since $Ha$ is a $1\times 1$ matrix). Now, if $n\ge 2$, the columns of $S$ are multiple of each other so $\det(B)=0$ and $f$ is degenerate.
We proved

If $f$ is non-degenerate, then $n=1$

If $n=1$, then $H=[m]$ and $h(a)=ma$ so $f(a,b)=m^2ab$. Clearly, $f$ is non-degenerate if and only if $m\neq 0$, which is equivalent to $h$ 1-1. As a conclusion,

$f$ is non degenerate if and only if $n=1$ and $h$ not the zero linear form.


Addendum:

Let $g$ be a 1-1 linear application from a vector space $V$ into a vector space $W$. Then, if $\{v_1,v_2,\dots,v_k\}$ is linearly independent, then $\{ g(v_1), g(v_2),\dots, g(v_k)\}$ is linearly independent. In particular, if $\mathcal B$ is a base of V, then $g(\mathcal B)$ is linearly independent.

Indeed, assume that $a_1g(v_1)+\dots a_kg(v_k)=0$, then $g(a_1v_1+\dots+a_kv_k)=0$. $g$ is 1-1 so $a_1v_1+\dots+a_kv_k=0$. But $\{v_1,v_2,\dots,v_k\}$ is linearly independent, so $a_1=a_2=\dots=a_k=0$. This means that $\{ g(v_1), g(v_2),\dots, g(v_k)\}$ is linearly independent.
As a consequence, let $\mathcal B=\{e_1,\dots,e_n\}$ be a basis of $M$. If $h$ is 1-1, then $\{h(e_1),\dots,h(e_n)\}$ should be linearly independent. But $\{ h(e_1), \dots, h(e_n)\}$ is a family of vectors in $\mathbb R$, so $n\le 1$.
