How to prove that $\phi(v\otimes f) = g(v,f)$ is injective?

Question

Let $V$ a finite vector space and $V^{\ast}$ its dual.

Let $g : V\times V^{\ast} \to \mathcal{L}(V,V)$ a bilinear map defined as follows:

$$g(v,f)(w) := f(w)v.$$

To show that the map $$\phi(v\otimes f) = g(v,f)$$ is injective, why have I to show that $\sum_{i}g(v_i,f_i) = 0 \Rightarrow \sum v_i\otimes f_i = 0?$

Answer 1

You have to because it shows $\ker\phi=0$. Indeed, by linearity, for a general tensor $$\phi\bigl(\sum_i v_i\otimes f_i\bigl)=\sum_i\phi(v_i\otimes f_i)\stackrel{\text{def}}{=}\sum_i g(v_i,f_i),$$ so that $$\sum_\limits ig(v_i,f_i)=0\iff \sum_i(v_i\otimes f_i)\in\ker\phi.$$

Answer 2

Write $x=\sum_{i\in I}v_i\otimes f_i$. Proof recursively on the on the cardinal of $I$. Suppose that the result is true if the cardinal of $I$ is $n$. Suppose that the cardinal of $I$ is $n+1$, $I=\{1,...,n+1\}$ $\phi(\sum_iv_i\otimes f_i)=0$, if $f_{n+1}=0$ done. Suppose $f_{n+1}\neq 0$, there exists $w$ such that $f_{n+1}(w)\neq 0$,

$\sum_if_i(w)v_i=0$ implies $v_{n+1}\in Vect(v_1,...,v_n)$ write $v_{n+1}=\sum_ia_iv_i, \sum_{i=1}^{i=n+1}v_i\otimes f_i=\sum_{i=1}^{i=n}v_i\otimes f_i+\sum_ia_iv_i\otimes f_i$

$\sum_{i=1}^{i=n}v_i\otimes (f_i+a_if_{n+1})=0$.

We apply the recursive hypothese to show that $\sum_{i=1}^{i=n+1}v_i\otimes f_i=\sum_{i=1}^{i=n}v_i\otimes (f_i+a_if_{n+1})=0$.

Answer 3

Comparing dimensions, it suffices you show your map is surjective, because both $V\otimes V^*$ and $\operatorname{End}(V,V)$ have dimension $(\dim V)^2$. Now consider any endomorphism $\eta : V\to V$, and pick a basis $(e_i)$ of $V$. Write $(\widehat {e_i})$ for the dual basis to $(e_i)$. Then any vector $v$ may be written as $$ v = \sum \widehat{e_i}(v) e_i$$ whence any linear endomorphism can be written as $$\eta(v) = \sum (\widehat{e_i}\circ \eta)(v) e_i=\sum g(e_i,\eta_i)(v)= g\left (\sum v_i\otimes \eta_i\right)$$ where $\eta_i = \widehat{e_i}\eta$.

Add Remark that basis play an essential role here. A claim that holds in general works for finitely generated projective modules, which admit finite dual bases. In this case, the dual basis we used is precisely $(e_i,\widehat{e_i})$. In general, a dual basis of an $A$-module $P$ is a collection of pairs $(x_i,\widehat{x_i})$ in $P\times P^\ast$ such that for every element $x\in P$ there are but finitely many $i$ with $\hat x_i(x)\neq 0$, and $$ x = \sum \widehat{x_i}(x) x_i.$$ For this one can check Bourbaki's algebra books, in the section where they discuss the relations between tensor products and hom functors.