Find signature and the symmetrical bilinear forms of $\phi(x^2) $ & $\phi(x)^2$ I have the following problem I struggle with :
Let $ \mathbb{K}  $ a commutative field [of different characteristics of $2$ ( it means that $1+1=2$ has an inverse $\in \mathbb{K}$)]. A $\mathbb{K}\text{-algebra}$ is a ring $E$ s.t it contains $\mathbb{K}$ as a subring ; It implies that $E$ is a vector space on $\mathbb{K}$ where the external operation is induced by the product on $E$. Let $\phi$ a linear form on $E$
We define :  $\;\;x  \rightarrow \phi(x)^{2}\;\:$ //&// $\;\: x \rightarrow \phi(x^{2})$
a) I have to proof that they're both quadratic forms and find their symmetric bilinear form
b) Then determine the signature for $x\rightarrow \phi(x)^{2}$ $(\mathbb{K}=\mathbb{R}$),
and for which condition over $\phi$ we can say that the q.f  $\;\phi(x^2)$ is equal to $(x,y)\rightarrow \phi(xy)$

The thing is that the function here is not defined , I'm used to work with function for exemple such as $\phi(x,y)= 3x^{2}+5xy+9y^2 $ anyway here's my thought on the problem but I think its all wrong:
a) We know that $\phi$ is a linear form so we can write it :
$\phi(x) = mx $ for $m \in \mathbb{K} \longrightarrow \phi(x)^2 = m^2x^2\;\;$  and $\;\:\phi(x^2) = mx^2$ is this enough to prove that they're both quadratic forms ? I still have no clue for their symmetric bilinear form
b) Using what I found above $ (\phi(x)^2 = m^2x^2) \; m^2 \geq 0$ hence the signature will be either positive or zero ? and no idea for the last part ..
Any help would be a lot appreciated. Thanks in advance for your help.
 A: Let $\mathbb{K}$ be a field of characteristic $\neq 2$, let $E$ be a $\mathbb{K}$-algebra, and fix some $\mathbb{K}$-linear map $\phi \colon E \to \mathbb{K}$. Then define the two functions
$$ Q_1(x) = \phi(x)^2 \quad \text{ and } \quad Q_2(x) = \phi(x^2). $$
In order to prove that these are both quadratic forms, it is enough to show that $Q_1(x) = B_1(x, x)$ for some bilinear form $B_1 \colon E \times E \to \mathbb{K}$, and similarly for $Q_2$. We can figure out what $B_1$ should be by polarising, so define
$$ \begin{aligned}
B_1(x, y)
  &= \frac{1}{2}(Q_1(x + y) - Q_1(x) - Q_1(y)) \\
  &= \frac{1}{2}(\phi(x + y)^2 - \phi(x)^2 - \phi(y)^2) \\
  &= \frac{1}{2}((\phi(x)+ \phi(y))^2 - \phi(x)^2 - \phi(y)^2) \\
  &= \phi(x) \phi(y).
\end{aligned}$$
Now it is easy to see that $B_1(x, y) := \phi(x) \phi(y)$ is bilinear, and $Q_1(x) = B_1(x, x)$, so $Q_1$ is indeed a quadratic form.
We can do the same thing for $Q_2$, arriving at $B_2(x, y) = \frac{1}{2} \phi(xy + yx)$ instead, which we can also verify (by the fact that $E$ is a $\mathbb{K}$-algebra which must have $\mathbb{K}$-bilinear multiplication) to be a bilinear map such that $B_2(x, x) = Q_2(x)$. Hence $Q_2$ is also a quadratic form. (If the $\mathbb{K}$-algebra $E$ is commutative, then we can simplify $B_2(x, y) = \phi(xy)$).
Now, suppose $\mathbb{K} = \mathbb{R}$. If $\phi = 0$, then both $Q_1$ and $Q_2$ are also zero, and so the form has signature $(0, \ldots, 0)$. Otherwise, suppose $\phi \neq 0$, and let $v \in E$ be a vector such that $\phi(v) \neq 0$. Defining $K = \ker \phi$, we get that $E = K \oplus \mathbb{R}v$ as a vector space, it is easy to see that $Q_1(v) = \phi(v)^2 > 0$, and that $Q_1(K) = 0$, so the signature is $(+, 0, \ldots, 0)$. Finally, I think that the signature of $Q_2$ can be any of  $(+, 0, \ldots, 0)$, $(0, 0, \ldots, 0)$, or $(-, 0, \ldots, 0)$, depending on the multiplication in $E$.
