Given two vector spaces $V$ and $W$, an *alternating multilinear operator* from $V^k$ to $W$ is a multilinear map

$ f: V^k \to X $

such that whenever $v_1,\cdots,v_k$ are linearly dependent vectors in $V$, then

$f(v_1,\ldots, v_k)=0$.

The *kth exterior power* of a linear space $V$, is a linear space $\Lambda^k(V)$ with alternating multilinear operator $\Lambda:V^k\to \Lambda^k(V)$ such that for any given linear space and any alternating multilinear operator $f:V^k\to W$, there is a unique linear map $\tilde f$from $\Lambda^k(V)$ to $W$ such that $\tilde f\circ \Lambda=f$ as indicated by the following commutative diagram:

\begin{xy}

\xymatrix{

V^k\ar[r]^{\Lambda} \ar[dr]_f & \Lambda^k(V)\ar[d]^{\tilde f}\\

& W

}

\end{xy}

**Construction**. Let $I$ be the linear subspace generated by $\{x_1\otimes\cdots\otimes x_k|x_1,\cdots,x_n \mbox{ are linear dependent}\}$ in $\otimes^k V$. Then $\otimes^k V/I$ is a kth exterior power.

*Proof*. Let $\bar f:\otimes^k(V)\to W$ be the linear map such that $\bar f\circ \

\otimes= f$, $q:\otimes^k(V) \to \otimes^k V/I$ be the quotient map, and $\tilde f([z])=\bar f(z)$.

\begin{xy}

\xymatrix{

V^k\ar[r]^{\otimes} \ar[dr]_f & \otimes^k(V)\ar[d]^{\bar f}\ar[r]^q & \otimes^k(V)/I\ar[dl]^{\tilde f}\\

& W

}

\end{xy}

If $z\in I$ then $z$ is a linear combination of element in $\{x_1\otimes\cdots\otimes x_k|x_1,\cdots,x_n \mbox{ are linear dependent}\}$, so $\tilde f([z])=\bar f(z)= \sum_{i=1}^n k_if(x_1^i,\cdots,x_n^i) = 0$, so $\tilde f$ is well-defined.

>Then what about the infinite exterior power?