## 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][1], 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.

## the atomic part of $C(X)$

Let $X$ be a compact Hausdorff space and $A=C(X)$. $\newcommand{\id}{\operatorname{id}}\newcommand{\C}{\mathbb{C}}$ For each state $\tau$ of $A$, dnote the associated GNS representation by $(\pi_\tau, H_\tau, x_\tau)$(or $(\pi_\tau,H_\tau)$). If $\tau$ is pure, then $\pi_\tau(A)’=\mathbb{C}\id_\tau$. Since $\pi_\tau(A)\subset \pi_\tau(A)’$, $\pi_\tau(A)=\mathbb{C}\id_\tau$ and thus $\pi_\tau(A)’=B(H_\tau)$. Hence $B(H_\tau)=\C\id_\tau$ and $H_\tau\cong \C$.
Suppose $\pi_\tau(a)=k\id_\tau$, then $k=\langle \pi_\tau(a)x_\tau,x_\tau\rangle=\tau(a)$, so $\pi_\tau(a)=\tau(a)\id_\tau$.

If $\tau$ and $\rho$ are two pure states of $A$, $(\pi_\tau, H_\tau)$ is unitarily equivalent to $(\pi_\rho, H_\rho)$. Then $\tau(a)\id_\tau=\pi_\tau(a)=u^*\pi_\rho(a) u=\rho(a)u^*\id_\rho u=\rho(a)\id_\tau$,where $u$ is a unitary from $H_\tau$ to $H_\rho$. Therefore, $\tau=\rho$. That is,
the atomic representation of $C(X)$ is $(\pi_{atomic}, H_{atomic})=(\oplus_{\tau\in PS(A)}\pi_\tau,\oplus_{\tau\in PS(A)} H_\tau)=(\oplus_{\tau\in \Sigma(A)}\tau(\cdot)\id_\tau, \oplus_{\tau\in \Sigma(A)}\C)=(\oplus_{\tau\in \Sigma(A)}\tau(\cdot)\id_\tau, \ell^2(\Sigma(A))),$ where $\Sigma(A)$ is the character space of $A$.

Since $\oplus_{\tau\in PS(A)}\tau(a)\id_\tau=\widehat{a}:\tau\to \tau(a)\id_\tau$, $\pi_{atomic}(A)=C(\Sigma(A))$. As $\pi_{atomic}$ is non-degenerate, operators in $B(\oplus_{\tau\in PS(A)}H_\tau)$ commuting with $\oplus_{\tau\in PS(A)}\pi_\tau(a)$ are of form $\oplus_{\tau\in PS(A)}u_\tau$. By $\|\oplus_{\tau\in PS(A)}u_\tau\|=\sup_{\tau\in PS(A)}\|u_\tau\|< \infty$ we know that $\pi_{atomic}(A)’=\ell^\infty(\Sigma(A))$ and thus $\pi_{atomic}(A)”=\ell^\infty(\Sigma(A))$.

## 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.