Notes:Dual basis

Note: this article is very pedantic. Which is why the concrete-to-abstract isomorphism for example is invoked.

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TODO: Finish off and turn into a task

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Basis

Let $(V,\mathcal{K})$ be a finite dimensional vector space over the field, $\mathcal{K}$, suppose it has dimension $n\in\mathbb{N}$.

• Let $E:=\{E_1,\ldots,E_n\}$ be any basis of $V$.
• Suppose $V^*$ is the set consisting of all functions, $f:V\rightarrow\mathcal{K}$ which are linear maps.
  • That is for $f\in V^*$ we have $f:V\rightarrow\mathcal{K}$ and:
    • $\forall \alpha,\beta\in\mathcal{K}\ \forall u,v\in V[f(\alpha u+\beta v)=\alpha f(u)+\beta f(v)]$ (the map is linear)

Then I claim:

• $\{\varepsilon_1,\ldots,\varepsilon_n\}$ is a basis of $V^*$ - the dual space to $V$ where we define:
  • $\varepsilon_i:V\rightarrow\mathcal{K}$ (obviously $\varepsilon_i\in V^*$) by $\varepsilon_i:=\varepsilon_i'\circ \Phi_E$.
    • $\Phi_E:V\rightarrow\mathcal{K}^n$ is the concrete-to-abstract isomorphism given by: $\Phi_E:v_1E_1+\cdots+v_nE_n\mapsto(v_1,\ldots,v_n)^T$ (maps to a column vector) and
    • $\varepsilon_i':\mathcal{K}^n\rightarrow\mathcal{K}$ defined by: $\varepsilon_i':(v_1,\ldots,v_{i-1},v_i,v_{i+1},\ldots,v_n)\mapsto v_i$
  • Thus $\varepsilon_i:\sum_{j=1}^n v_jE_j\mapsto v_i$

Proof

The proof is done in 2 parts, first we must show that $\text{Span}$$(\{\varepsilon_1,\ldots,\varepsilon_n\})=V^*$, then that the set of (co)vectors $\{\varepsilon_1,\ldots,\varepsilon_n\}$ are a linearly independent set.

Span part

• Let $f\in V^*$ be given. We will show this implies $f\in\text{Span}(\{\varepsilon_1,\ldots,\varepsilon_n\})$, and thus that $\text{Span}(\{\varepsilon_1,\ldots,\varepsilon_n\})\subseteq V^*$
  • First note that we can say $f=g\iff \forall v\in V[f(v)=g(v)]$ so rather than considering trying to show $(a,b)\in f\iff (a,b)\in g$ as a relation we can instead deal with it as a map.
  • Let $v\in V$ be given. Note that we can write $v=\sum_{j=1}^n v_jE_j$ which we shall write as $\sum v_jE_j$ for short on this page.
    • Now $f(v)=f(\sum v_jE_j)=\sum v_jf(E_j)$ by linarity of $f$.
      • Note that $f(\alpha E_j)=k_j\varepsilon_j(\alpha E_j)$ for some $k_j\in\mathcal{K}$. Choose $k_j:=\frac{f(E_j)}{\varepsilon_j(E_j)}$ and the result follows. Thus:
      • we see $f(v)=f(\sum v_jE_j)=\sum v_jf(E_j)=\sum v_jk_j\varepsilon_j(E_j)=\sum (v_jk_j)\varepsilon_j(E_j)$
  • Since $v\in V$ was arbitrary we have shown: $\forall v\in V[f(v)=\left(\sum (\cdot k_j)\varepsilon_j(E_j)\right)(v)]$
  • Thus we see $f=\left(\sum (\cdot k_j)\varepsilon_j(E_j)\right)$
• Thus $f\in\text{Span}(\{\varepsilon_1,\ldots,\varepsilon_n\})$

Going the other way, to show that $\text{Span}(\{\varepsilon_1,\ldots,\varepsilon_n\})\subseteq V^*$ is trivial. In fact as we know already $V^*$ is a vector space we might be able to show it just by applying what we already know! Either way, it's not hard.

Linear independence part

• Now we want to show that $\{\varepsilon_1,\ldots,\varepsilon_n\}$ is a linearly independent set. This means we want to show:
  • $\left(\left(\sum \alpha_i\varepsilon_i\right)=\underline{0}\right)\implies\left(\forall i\in\{1,\ldots,n\}[\alpha_i=0]\right)$, here $\underline{0}$ denotes the map $\underline{0}:V\rightarrow\mathbb{K}$ by $\underline{0}:v\mapsto 0$, lets prove this:
• Let $i\in\{1,\ldots,n\}$ be given
  • Suppose $\alpha_i\ne 0$
    • Choose $v=0E_1+\cdots+OE_{i-1}+v_iE_i+0E_{i+1}+\cdots 0E_n$ (which perhaps can be written better as: $(0,\ldots,0,v_i,0,\ldots,0)$ where $v_i\ne 0$
      • Then by hypothesis: $\left(\sum \alpha_j\varepsilon_j\right)(v)=0$, so:
        • $\sum \alpha_j\varepsilon_j(v)=\sum \alpha_j\varepsilon_j(\sum_{k=1}^n v_iE^i)$ blah blah blah
      • We arrive at: $\sum \alpha_j\varepsilon_j(v)=\alpha_iv_i$, but as the LHS $=0$ we see we have:
        • $\alpha_iv_i=0$
        • We know $v_i\ne 0$, so $\alpha_iv_i=0$ means we must have $\alpha_i=0$. This contradicts that $\alpha_i\ne 0$, so we must have $\alpha_i=0$
• Since $i$ was arbitrary we see this is true for all alpha