I have a question from my professor' notes. We defined $\Lambda^k(V)$ as the set of all $k$-Tensor' forms (multilinear transformations), $\omega$, which fulfuill $\omega(v_1,...,v_i,v_j,...,v_k)=-\omega(v_1,...,v_j,v_i,...,v_k)$.
Let $v_1,...,v_{n-1}\in\mathbb{R}^n$. Let $\varphi:\mathbb{R}^n\to\mathbb{R}$ defined by $\varphi(\omega)=\operatorname{det}\left(\begin{smallmatrix} v_1\\ \vdots\\ v_{n-1}\\ \omega \end{smallmatrix}\right)$. We observe that $\varphi\in\Lambda^1(\mathbb{R}^n)$, thus, there exsists a unique $z\in\mathbb{R}^n$ such that $$\\ \left \langle z,\omega \right \rangle=\varphi(\omega)=\operatorname{det}\left(\begin{smallmatrix} v_1\\ \vdots \\ v_{n-1}\\ \omega \end{smallmatrix}\right) \ $$
My question is why is there a uniuqe $z$ as above?
I know that $\varphi(\omega)$ is some number in $\mathbb{R}$, so I have a lot of options to choose $z$. For example, if $\varphi(\omega)=3$ and $\omega=(2,5,-24)$ and $z=(z_1,z_2,z_3)$, I can find a lot of options such that $2z_1+5z_2-24z_3=3$, so how is $z$ unique?