Exterior power of vector space

Question

Let $V$ be an n-dim. $\mathcal{F}$-vector space. Then I guess we have $\Lambda^k[V^*]\cong \{\text{skew-symmetric multilinear maps } V^k\to \mathcal{F}\}$ but I don't recall what the isomorphism is given by.

Is it $y_1\wedge \cdots \wedge y_k \mapsto \left( (v_1,\cdots, v_k)\mapsto \prod_iy_i(v_i) \right)$ ??

I just don't see why that's skew-symmetric.

Answer 1

The isomorphism is given by

$$y_1 \wedge \dots \wedge y_k (v_1,\dots,v_k) = \begin{vmatrix}y_1(v_1) & \cdots & y_1(v_k) \\ \vdots & & \vdots \\ y_k(v_1) & \cdots & y_k(v_k) \end{vmatrix}$$

It is skew-symmetric due to the properties of determinant.

Picking up a basis for $V$ and a dual one for $V^*$ shows that it is an isomorphism.