True/False question on a bilinear map Let $A=\begin{bmatrix}1&2\\ 4&3 \end{bmatrix}$. Let $\phi:\mathbb{R}^2\times\mathbb{R}^2\to\mathbb{R}$ be the bilinear map defined by  $\phi(v,w)=v^{T}Aw$, 
then  choose the correct statement
My answer : option 1) will true by taking $B =\begin{pmatrix}1&0\\0&\!-1\end{pmatrix}$. Indeed: 
$$\phi(v,v)=(v_1\;v_2)\begin{pmatrix}1&0\\0&\!-1\end{pmatrix}\binom{v_1}{v_2}=(v_1\;-v_2)\binom{v_1}{v_2}=v_1^2-v_2^2\implies v_1=\pm v_2$$
I'm confused about option $2)$. Any hint/solution will be appreciated. Thank you.
 A: Let $\Bbb F$ be any field with $\text{char}(\Bbb F) \ne 2$, and let
$A \in M_n(\Bbb F) \tag 1$
be any $n \times n$ matrix with entries chosen from $\Bbb F$; define the $\Bbb F$-bilinear form
$\phi_A: \Bbb F^n \times \Bbb F^n \to \Bbb F; \; \phi_A(x, y) = x^T A y \in \Bbb F, \; x, y \in \Bbb F^n; \tag 2$
as is well-known, we may define the matrices
$A_\sigma = \dfrac{1}{2}(A + A^T), \tag 3$
$A_\alpha = \dfrac{1}{2}(A - A^T); \tag 4$
we have
$A_\sigma^T = \dfrac{1}{2}(A + A^T)^T = \dfrac{1}{2}(A^T + (A^T)^T) = \dfrac{1}{2}(A + A^T) = A_\sigma, \tag 5$
$A_\alpha^T = \dfrac{1}{2}(A - A^T)^T = \dfrac{1}{2}(A^T - (A^T)^T) = \dfrac{1}{2}(A^T - A) = -\dfrac{1}{2}(A - A^T) = -A_\alpha; \tag 6$
also,
$A = \dfrac{1}{2}(A + A^T) + \dfrac{1}{2}(A - A^T) = A_\sigma + A_\alpha; \tag 7$
$A_\sigma$ and $A_\alpha$ are respectively the $\sigma$ymmetric and $\alpha$ntisymmetric parts of $A$; it follows then that
$\phi_A(x, y) = x^TAy = x^T(A_\sigma + A_\alpha)y = x^T(A_\sigma y + A_\alpha y) = x^TA_\sigma y + x^T A_\alpha y; \tag 8$
taking $y = x$ we have
$\phi_A(x, x) = x^T A_\sigma x + x^T A_\alpha x; \tag 9$
now,
$x^T A_\alpha x \in \Bbb F, \tag{10}$,
and thus
$x^T A_\alpha x = (x^T A_\alpha x)^T = x^T A_\alpha^T (x^T)^T = x^T(-A_\alpha) x = -x^T A_\alpha x, \tag{11}$
whence
$2x^T A_\alpha x = 0; \tag{12}$
since $\text{char}(\Bbb F) \ne 2$ we see that
$x^T A_\alpha x = 0; \tag{13}$
it then follows from (9) that
$\phi_A(x, x) = x^T A_\sigma x; \tag{14}$
we see that $\phi(x, x)$ is determined solely by the $\sigma$ymmetric part of $A$.
Taking $\Bbb F = \Bbb R$ and
$A = \begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix}, \tag{15}$
we see that with 
$B = A_\sigma = \dfrac{1}{2} (A + A^T) = \dfrac{1}{2} \left ( \begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix} + \begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix} \right ) = \dfrac{1}{2}\begin{bmatrix} 2 & 6 \\ 6 & 6 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 3 & 3 \end{bmatrix} = B^T, \tag{16}$
that
$\phi(x, x) = x^TBx, \tag{17}$
and thus that statement (1) is correct.
Statement (2) is false, since
$\psi \left ( \begin{bmatrix} \beta v_1 \\ \beta v_2 \\ \beta w_1 \\ \beta w_2 \end{bmatrix} \right ) = \phi \left ( \begin{bmatrix} \beta v_1 \\ \beta v_2 \end{bmatrix}, \begin{bmatrix} \beta w_1 \\ \beta w_2 \end{bmatrix} \right ) = \beta^2 \phi \left ( \begin{bmatrix}  v_1 \\  v_2 \end{bmatrix}, \begin{bmatrix} w_1 \\ w_2 \end{bmatrix} \right ) =  \beta^2 \psi \left ( \begin{bmatrix}  v_1 \\  v_2 \\  w_1 \\ \ w_2 \end{bmatrix} \right ), \tag{18}$
so $\psi$ cannot be linear.
Finally, in response to a remark made by our OP Messi fifa, I'm not sure how the matrix
$B = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \tag{19}$
relates to the problem at hand; $\phi_A(x, x) = x^TBx$ only holds for certain $B$, e.g. $B = A_\sigma$; e.g, it is clear with $x = (1, 1)^T$ that $x^TBx = 0 \ne x^TAx = 10$.
