Q: Suppose $\{e_1,\ldots,e_n\}$ is a basis of a vector space V over the field F . Let $e^1,\ldots,e^n\in$ $V^*$ be the linear functionals defined by $$ e^i(e_j) = \begin{cases} c, i=j \\
0, i\neq j \end{cases}$$ Then prove that the set $\{e^1,\ldots,e^n \}$ is linearly intependent .
My approach:
$\forall\,v:=(x_1,\ldots,x_n)\in V,a_1e^1(v)+\ldots+a_ne^n(v)=0$
In order to prove $\{e^1,\ldots,e^n \}$ is linearly independent I have to show that $a_1=\ldots=a_n=0$ . But here I confused how to show that . any hint or solution will be appreciated .
Extra question: I really confused why $e^1,\ldots,e^n$ defined like that $?$ Is it for make $e^i$ unique or something else .
I am newly introduce with dual space so sorry if my question is rudimentary.
Thanks in advance .
Prove that the dual basis is linearly independent
dual-spaceslinear algebra
Best Answer
The notation is somewhat unusual, but the $e^i$ are the coordinate functions for the basis $e_1,\ldots,e_n$ of $V$. That is, from any linear combination $x_1e_1+\cdots+x_ne_n$, where the $x_1,\ldots,x_n$ are arbitrary scalars, the function $e^j$ extracts the coordinate $x_j$: $$ e^j(x_1e_1+\cdots+x_ne_n)=x_1e^j(e_1)+\cdots+x_ne^j(e_n)\\=x_1\cdot0+x_2\cdot0+\cdots+x_j\cdot1+\cdots+x_n\cdot0=x_j. $$ Now to the question itself: here the linear combination is over the coordinate functions, so the result is an arbitrary linear function on $V$ with scalar values. To show that this is the null function only if all the coefficients are zero, you need to extract an individual coefficient $a_i$. The trick is quite similar to the above, only now it is not applying $e^j$, but applying to $e_i$ what is needed: $$ (a_1e^1+\cdots+a_ne^n)(e_i) =a_1e^1(e_i)+\cdots+a_ne^n(e_i)\\ =a_1\cdot0+\cdots+a_i\cdot1+\cdots+a_n\cdot0 =a_i. $$ So it the linear form in the leftmost parentheses is zero, then certainly $a_i=0$, and this holds for every $i$.