# A basis for the dual space of $V$

We let $\mathcal{B} = \{\alpha_1, ... \alpha_n\}$ be a basis for $V$, a vector space with inner product $\langle \cdot, \cdot \rangle$. Then we define $f_i(v) := \langle v, \alpha_i \rangle$.

Show that $f_1, ..., f_n$ is a basis for $V^{\star}$ the dual space of $V$.

This was a question on a midterm I just finished, I honestly had no idea how to do it here's my relevant work:

I know that if $f_1, ..., f_n$ are independent than they are a basis since $\dim(V^{\star}) = \dim(V) = n$ because they are finite dimensional. So I have to show that they are independent, this is where I am totally lost.

I can show that if $\mathcal{B}$ is orthogonal then they are linearly independent, since if $c_1f_1 + ... + c_nf_n = 0$ implies we must have $(c_1f_1 + ... + c_nf_n)(\alpha_i) = 0$ for each $\alpha_i$ and since the $\alpha_i$ are orthogonal then for each $i$ this becomes: $c_i \langle \alpha_i, \alpha_i \rangle = 0$ but $\langle \alpha_i, \alpha_i \rangle \neq 0$ since $\alpha_i \neq 0$ and hence $c_i = 0$ for all $i$.

Now I wonder is there a way to generalize an approach like this to prove the actual question posed here, or am I going about this completely wrong? If the second is the case, how should I have gone about solving this problem?

Thanks,

Suppose the $f_i$ are dependent, so $\sum_i c_i f_i = 0$ for some scalars $c_i$. What does it mean for an element of $V^*$ to be zero? Well, such elements are defined to be linear functionals on $V$ and a function is zero if it takes the value zero everywhere on its domain. So for all $v\in V$ we have $0 = \left(\sum_i c_i f_i\right)(v) = \sum_i c_i f_i(v) = \sum_i c_i\langle v,\alpha_i\rangle = \left\langle v,\sum_ic_i\alpha_i\right\rangle,$ where the first equation is the definition of a function being zero, the second is the definition of the vector space operations on $V^*$, the third is the definition of $f_i$, and the fourth is linearity of $\langle\cdot,\cdot\rangle$ in the second coordinate.
In particular, we can take $v = \sum_i c_i\alpha_i$, so we obtain $\left\langle \sum_i c_i\alpha_i,\sum_i c_i\alpha_i\right\rangle=0$. The positive definiteness property of an inner product says that the only $v$ with $\langle v,v\rangle=0$ is $v=0$, so $\sum_i c_i\alpha_i = 0$.
By assumption $\{\alpha_1,\ldots,\alpha_n\}$ is a basis for $V$, so it is linearly independent. Therefore $\sum_i c_i\alpha_i=0$ implies $c_i = 0$ for all $i$.
Since we started with the assumption that $\sum c_i f_i = 0$ and ended up with $c_i = 0$ for all $i$, we have shown that the set $\{f_1,\ldots,f_n\}$ is linearly independent as well. Since you know $\dim(V) = \dim(V^*) = n$, this means that $\{f_1,\ldots,f_n\}$ is a basis for $V^*$.