## linear dependence of linear functionals

**Lemma**. Suppose $\tau_1,\cdots,\tau_n$ are linear functionals on a linear space $V$, then they are linear dependent if and only if
\begin{align*}
\det(\tau_i(v_j))=0 \forall v_1,\cdots, v_n\in V.
\end{align*}

*Proof*. We need only prove the ONLY IF part.
If $\tau_1,\cdots,\tau_n$ are linearly independent, by [this post], there is some $v\in V$ such that
$$v\in (\bigcap_{i=1}^{n-1}\ker\tau_i)\backslash \ker \tau_n.$$
$\newcommand{diag}{\operatorname{diag}}$
Fix $v_n=v$, then $$\det(\tau_i(v_j))_{1\leq i,j\leq n-1}=\tau_n(v_n)^{-1}\det(\tau_i(v_j))_{1\leq i,j\leq n}=0.$$
Apply induction.

## the Exterior Power of Dual Spaces

We have known that the exterior power of vector spaces exists. We give a more convenient construction for finite dual spaces.

Let $\tilde \pi$ be filled in the commutative diagrams:
\begin{xy}
\xymatrix{
\Pi^n V^* \ar[r]^\wedge \ar[rd]_\pi & \wedge^n V^* \ar@{–>}[d]^{\tilde\pi}\\
& (\Pi^n V)^*,
}
\end{xy}
where $\pi$ is an alternating operator defined by $$\pi(v’_1,\cdots,v’_n):(v_1,\cdots,v_n)\mapsto \det(v’_i(v_j)).$$
We will show $\tilde\pi$ is an injection.

## Equivalent condition of linear dependence of linear functionals

**Theorem**. Suppose $f_1,\cdots, f_n$ and $g$ are linear functional on a vector space. Then $g$ is linearly spanned by $f_1,\cdots,f_n$ if and only if $$\bigcap_{i=1}^{n}\ker f_i\subset \ker g.$$

*Proof*. We prove this by induction. If $n=1$ and $g\neq 0$, then there is some $x_1\notin \ker f_1$. Since $$f_1(x)x_1-f_1(x_1)x\in \ker f_1\subset \ker g,$$
$f_1(x)g(x_1)-f_1(x_1)g(x)=0,$
i.e., $$g(x)=\frac{g(x_1)}{f_1(x_1)}f_1(x).$$

Assume the proposition is true for the case of $n-1$, consider the case of $n$.

If there is some $i_0\in \{1,\cdots,n\}$ such that $\cap_{i\neq i_0}\ker f_i\subset\ker f_{i_0},$ then $\cap_{i\neq i_0}\ker f_i=\cap_{i=1}^{n}\ker f_i\subset \ker g$, then the proposition holds for the case of $n$ by assumption.

If for any $j\in \{1,\cdots,n\}$ there is some $x_j\in (\cap_{i\neq j}\ker f_i)\backslash\ker f_j$, then $$x-\sum_{j=1}^n\frac{x_j}{f_j(x_j)}f_j(x)\in \cap_{i=1}^n\ker f_i.$$
so $g(x-\sum_{j=1}^n\frac{x_j}{f_j(x_j)}f_j(x))=0$, i.e., $g(x)=\sum_{j=1}^n\frac{g(x_j)}{f_j(x_j)}f_j(x)$ for all $x$.

Hence the proposition holds for the case of $n$.

## Exterior Power

**Definition**. Suppose $I$ is an index set, $V$ and $W$ are linear vector spaces, a multilinear map $f:\oplus_{i\in I} V\to W$ is called an *alternating operator* if $f((v_i)_{i\in I})=0$ whenever $(v_i)_{i\in I}\in \oplus_{i\in I} V$ is a family of linearly dependent vectors in $V$, here $\oplus_{i\in I} V$ merely means a subspace of $\Pi_{i\in I} V$. The I-*exterior power* of a linear space $V$ with a linear space $\Lambda^I V$ with an alternating operator $\Lambda$ such that for any linear space $W$ and an alternating operator $f$, there exists a unique linear map $\tilde f$ such that the following diagram is commutative:
\begin{xy}
\xymatrix{
\oplus_{i\in I} V\ar[r]^{\Lambda} \ar[rd]_f&\Lambda^I V\ar@{–>}[d]^{\tilde f}\\
& W
}
\end{xy}

**Existence**. Let $L$ be the linear subspace of $\otimes^I V$ gernerated by $\{v_1\otimes\cdots \otimes v_n |v_1,\cdots, v_n \mbox{ are linearly dependent. }\}$, then $\otimes^I V/L$ with $q\circ \otimes$ (where $q$ is the quotient map) is the I-*exterior power* of $V$.

*Proof*. For any alternating operator $f:V\to W$, we know that there exists some linear map $\bar f$ s.t. the left triangle commutative in the following diagram. Define $\tilde f$ as the map induced by $\bar f$, i.e., $$\tilde f(q(x))=\bar f(x)\forall x\in \otimes^I V.$$
\begin{xy}
\xymatrix{
\oplus_{i\in I} V\ar[r]^{\otimes} \ar[rd]_f&\otimes^I V\ar@{–>}[d]^{\bar f}\ar[r]^q&\Lambda^I V\ar@{–>}[dl]^{\tilde f}\\
& W
}
\end{xy}
Since the image of $\tilde f$ is deterministic on each generator, $\tilde f$ is deterministic.

## Tensor Algebra

**Definition**. Suppose $V$ is a vector space, the tensor algebra of $V$ is defined as an algebra $TV$ with a linear map $T:V\to TV$ such that for any algebra $A$ and any linear map $f:V\to A$ there exists a unique algebra homomorphism $\tilde f:TV\to A$ such that $f=\tilde f\circ T$ as indicated by the commutative diagram below.
\begin{xy}
\xymatrix{
V\ar[r]^T\ar[rd]_f & TV\ar@{–>}[d]^{\tilde f}\\
& A }
\end{xy}

**Existence**. $TV$ is defined as $\oplus_{n\in \mathbb N}\otimes^n V$ and $T$ is defined as a map send a vecctor $v$ to an assignment of $v$ to 1 and 0 to others.

*Proof*. Since $\tilde f\circ T(v)=f(v)$, $\tilde f(\otimes_{j}v_j)=\Pi_j f(v_j)$ where $\otimes_{j}v_j$ is an element of $\otimes^n V_j$, $f(\oplus_{i} w_i)=f(\sum_i w_i\chi_i)=\sum_i f(w_i)$. Hence $\tilde f$ is unique if it exists.

Let $\tilde f_i$ be the uqnique linear map as indicated by the following commutative diagram:

\begin{xy}
\xymatrix{
\Pi^i V\ar[r]^\otimes\ar[rd]_{\Pi^i f} & \otimes^i V\ar@{–>}[d]^{\tilde f_i}\\
&W}
\end{xy}
Define $\tilde f=\sum_i \tilde f_i .$ We only have to prove $\tilde f$ preserves product. $\tilde f((\oplus_n u_n)(\oplus_n v_n))=\tilde f(\oplus_n \sum_{m+l=n} u_m\otimes v_l)=\sum_n \sum_{m+l=n}\tilde f_n(u_m\otimes v_l)=\sum_n\sum_{m+l=n} \tilde f_m(u_m) \tilde f_l(u_l)=\sum_m \tilde f_m(u_m)\sum_l \tilde f_l(v_l)=\tilde f(\oplus_n u_n)\tilde f(\oplus_n v_n).$