vProve that the $\phi's$ form a basis for $V^*$

Question

Hey guys. I've learned linear algebra before, but I kinda forget the part about dual space. For this problem, I think that because $V^*$ has the same dimension as V, which is n in this problem. And because the number of $\phi's$ are also n. I only need to prove that they are linearly independent, to prove that it is a dual basis? Thanks!

Also if I know that ${(3,1),(2,-2)}$ is a basis for $\Bbb R^2$, how do I find the dual basis for $(\Bbb R^2)^*$?

Answer 1

The dual basis of a basis $\mathcal B=(e_1,\dots, e_n)$ is the basis of linear forms defined by $$e_i^*(e_j)=\delta_{ij}~(\text{Kronecker's symbol})=\begin{cases}1&\text{if }i=j,\\0&\text{if }i\ne j.\end{cases}$$ This characterisation is enough to prove the linear independence of $\mathcal B'=(e_1^*, \dots, e_n^*)$.

Indeed, if $\lambda_1e_1^*+\dots+\lambda_n e_n^*=0$, then for each $i=1,\dots,n$, $$(\lambda_1e_1^*+\dots+\lambda_n e_n^*)(e_i)=\lambda_i=0.$$

Answer 2

$V$ is a vector space and $\mathcal{B}$ is a basis for $V$, then an arbitrary function $f: \mathcal{B}\rightarrow \mathbb{F}$ is a function, this function can be extended to a unique linear function $f\in{\rm Hom}_\mathbb{F}(V, \mathbb{F}).$

Answer 3

You can treat elements of $(\mathbb R^{2})^*$ as row vectors, so if $B$ is a matrix with the elements of a basis of $\mathbb R^2$ as its columns, the rows of $B^{-1}$ are the elements of the dual basis.