I am trying to understand the dual spaces and working on some problems.
Let $V=\mathbb{R}^{3} $ with the base $b=\left \{ (1,0,1),(1,2,1),(0,0,1) \right \}$ find the dual basis $ b^{\ast }$ of the dual space $ V^{\ast }$
let $ b^{\ast }=\left \{ f_1,f_2,f_3 \right \}$
i did some standar calculations and found that
$ f_1(x,y,z)=x-\frac{y}{2}$
$f_2(x,y,z)=\frac{y}{2}$
$f_3(x,y,z)=-x+z$
Could someone tell what this exactly means? who is the dual space $V^{\ast }$ in the end? Which function takes us from $V$ to $V^{\ast }$
I hope my questions have a meaning because i am a little confused at the moment.
The dual space $V^*$ is the set of all linear maps $V\to\Bbb R$, endowed with the usual (=pointwise) addition and scalar multiplication.
A specific isomorphism $V\to V^*$ through the given example can be given, specifically map the basis elements $u_1,u_2,u_3$ in $b$ to the basis elements $f_1,f_2,f_3$ in $b^*$. More specifically, $$(\alpha u_1+\beta u_2+\gamma u_3)\ \mapsto \ (\alpha f_1+\beta f_2+\gamma f_3)\,.$$
Nevertheless, it's easier to establish an isomorphism through the standard basis and its dual, and actually the standard inner product may also come handy, as using that, the standard isomorphism can simply be written as: $$v\mapsto\ (w\mapsto v^Tw)\,.$$