## Proof That a Real Valued Linear, Symmetric Function of Two Vectors in R3 Equals the Determinant of a f a 3x3 Matrix of the Two Vectors and a Third Vector

Theorem
Suppose
$\omega$
is a real valued function of vectors
$\mathbf{a}, \: \mathbf{b} \in \mathbb{R}^3$
such that
$\omega( \mathbf{a}, \mathbf{b})= \omega( \mathbf{b}, \mathbf{a})$
and that
$\omega$
is linear in both arguments.
There is a vector
$\mathbf{c}$
such that
$\omega( \mathbf{a}, \mathbf{b})= det (\mathbf{a} , \mathbf{b}, \mathbf{c})$
for all vectors
$\mathbf{a}, \: \mathbf{b}$
.
Proof
Let
$\mathbf{e}_1=(1,0,0)^T, \: \mathbf{e}_2=(0,1,0)^T, \: \mathbf{e}_3=(0,0,1)^T$
be the basis for
$\mathbb{R}^3$
.
Then we can write
$\mathbf{a}=a_1 \mathbf{e}_1 +a_2 \mathbf{e}_2 + a_3 \mathbf{e}_3$

$\mathbf{b}=b_1 \mathbf{e}_1 +b_2 \mathbf{e}_2 + b_3 \mathbf{e}_3$

$\omega$
is linear and symmetric in
$\mathbf{a} \: \mathbf{b}$
so
\begin{aligned} \omega(\mathbf{a}, \mathbf{b}) &= \omega(a_1 \mathbf{e}_1 +a_2 \mathbf{e}_2 + a_3 \mathbf{e}_3 ,b_1 \mathbf{e}_1 +b_2 \mathbf{e}_2 + b_3 \mathbf{e}_3 ) \\ &= a_1 b_2 \omega(\mathbf{e}_1 , \mathbf{e}_2) + a_1 b_3 \omega(\mathbf{e}_1 , \mathbf{e}_3)+a_2 b_1 \omega(\mathbf{e}_2 , \mathbf{e}_1) \\ &+ a_2 b_3 \omega(\mathbf{e}_2 , \mathbf{e}_3) + a_3 b_1 \omega(\mathbf{e}_3 , \mathbf{e}_1) + a_3 b_2 \omega(\mathbf{e}_3 , \mathbf{e}_2) \\ &= (a_2 b_3-a_3 b_2) \omega(\mathbf{e}_2, \mathbf{e}_3), + (a_3 b_1-a_1b_3) \omega(\mathbf{e}_3, \mathbf{e}_1) \\ &+ (a_1 b_2-a_2b_1) \omega(\mathbf{e}_1, \mathbf{e}_2), \end{aligned}
(1)
Let
$\omega(\mathbf{e}_2 , \mathbf{e}_3) =c_1, \: \omega(\mathbf{e}_3 , \mathbf{e}_1) =c_2 , \: \omega(\mathbf{e}_1 , \mathbf{e}_2) =c_3$

With these substitutions we can write (1) as
\begin{aligned} & (a_2 b_3-a_3b_2) c_1 + (a_3 b_1-a_1b_3) c_2 + (a_1 b_2-a_2b_1) c_3 \\ &= det \left( \begin{array}{ccc} a_1 &b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array} \right) \\ &=det (\mathbf{a} , \mathbf{b}, \mathbf{c}) \end{aligned}