# Prove that $\Omega(x^{*^1}, \ldots, x^{*^n}; x_1, \ldots, x_n) = \phi (x^{*^1}, \ldots, x^{*^n}) \Delta(x_1, \ldots, x_n)$

In the book Linear Algebra by Werner Greub, at page $112$, it is given that,

Let $E^*, E$ be a pair of dual vector spaces and $\Delta^* \not = 0, \Delta \not = 0$ be determinant functions in $E^*$ and $E$. It will be shown that $$\Delta^*(x^{*1}, \ldots, x^{*n}) \Delta(x_1, \ldots, x_n) = \alpha \det(\langle x^{*i}, x_j \rangle), \quad x^{*i} \in E^*, x_j \in E,$$ where $\alpha$ is a scalar.

Consider the function $\Omega$ of $2n$ vectors defined by $$\Omega(x^{*1}, \ldots, x^{*n}; x_1, \ldots, x_n) = \det(\langle x^{*i}, x_j \rangle).$$

Then it follows from the properties of the determinant of a matrix that $\Omega$ is linear with respect to each argument. Moreover, $\Omega$ is skew symmetric with respect to the vectors $x^{*i}$ and with respect to the vectors $x_i$ (i= 1...n). Hence the uniqueness theorem (sec. 4.3) implies that $\Omega$ can be written as $$\Omega(x^{*1}, \ldots, x^{*n}; x_1, \ldots, x_n) = \phi (x^{*1}, \ldots, x^{*n}) \Delta(x_1, \ldots, x_n)$$

However, the uniqueness theorem is valid(from the its statement) only if $\Omega$ take only one of the $n$ vectors as its input, i.e

So how can we prove that $\Omega(x^{*1}, \ldots, x^{*n}; x_1, \ldots, x_n)$ can be written as $$\Omega(x^{*1}, \ldots, x^{*n}; x_1, \ldots, x_n) = \phi (x^{*1}, \ldots, x^{*n}) \Delta(x_1, \ldots, x_n)$$

The author is arguing as follows: for any fixed $n$-tuple $(x^{*1},\ldots,x^{*n})$ of covectors, the function $$(x_1,\ldots,x_n) \mapsto \Omega(x^{*1},\ldots,x^{*n};x_1,\ldots,x_n)$$ is multilinear and skew-symmetric (i.e., it is a determinant function). Therefore by the uniqueness theorem of determinant functions, there exists a scalar $\Phi(x^{*1},\ldots,x^{*n})$ (depending on our fixed $n$-tuple) such that our determinant function is just a multiple of the fixed non-trivial determinant function $\Delta$, i.e., $$\Omega(x^{*1},\ldots,x^{*n};x_1,\ldots,x_n) = \Phi(x^{*1},\ldots,x^{*n}) \Delta(x_1,\ldots,x_n).$$
Stated differently: the set of all determinant functions on $E$ actually forms a vector space. (If you're familiar with exterior algebra, it is the space $(\Lambda ^n E)^*$.) The uniqueness theorem above amounts to the statement that this space is one-dimensional, spanned by $\Delta$. Therefore there is an isomorphism $\phi:(\Lambda ^n E)^* \to K$ characterized by $\phi(\Delta) = 1$, where $K$ denotes the base field. The author is implicitly defining a map which I will call $\psi:(E^*)^n \to (\Lambda ^n E)^*$, by the rule $$\psi(x^{*1},\ldots,x^{*n}) = [(x_1,\ldots,x_n) \mapsto \Omega(x^{*1},\ldots,x^{*n};x_1,\ldots,x_n)].$$ Then his $\Phi$ is precisely the composition $\Phi = \phi \circ \psi$.
• Do you mean $\bigwedge$ ("\bigwedge") or $\Lambda$ ("\Lambda")? – Eric Towers Aug 4 '17 at 21:14
• @onurcanbektas It maps the $n$-tuple of covectors $(x^{*1},\ldots,x^{*n})$ to the determinant function that maps the $n$-tuple of vectors $(x_1,\ldots,x_n)$ to $\Omega(x^{*1},\ldots,x^{*n};x_1,\ldots,x_n)$. – Alex Provost Aug 5 '17 at 5:55