Suppose:
- $V$ and $W$ are two finite-dimensional vector spaces over $\mathbb{R}$
- $\dim(V)=n$, $\dim(W)=m$
- $\mathcal{E}=\{e_i\}_1^n\in V^n$ is an ordered basis for $V$
- $e_i^*:V\to\mathbb{R}$ is the $i$-th coordinate functional with regard to $\mathcal{E}$, thus $\forall v\in V,\ v=\sum_i^n e_i^*(v)\cdot e_i$
- $\mathcal{E}^*=\{e_i^*\}_1^n$ is the dual basis for $V^*$ (the dual space of $V$).
I feel it's difficult to grasp the concept of dual basis, although $\forall f\in \hom(V,\mathbb{R})$, the following identity suggests so:
$$f(v)=f\left(\sum_{i=1}^n e_i^*(v)\cdot e_i\right)=\sum_{i=1}^n e_i^*(v)\cdot f(e_i)=\sum_{i=1}^n f(e_i)\cdot e_i^*(v)$$
In the identity above, the change of the order of $e_i^*(v)$ and $f(e_i)$ is interesting.
For $\tau\in\hom(V,W)$, we have a similar identity, wrt $\mathcal{E}$:
$$\tau(v)=\tau\left(\sum_{i=1}^n e_i^*(v)\cdot e_i\right)=\sum_{i=1}^n e_i^*(v)\cdot \tau(e_i)$$ But I don't know if there is also a meaningful interpretation if the order of the two terms are swapped:
$$\tau(v)=\sum_{i=1}^n \tau(e_i)\cdot e_i^*(v)$$ It seems that the skeleton of $f$, i.e., $\{f(e_i)\}_1^n$ happen to be in $\mathbb{R}$, thus they can treated as coefficients (so the change of order); but the skeleton of $\tau$, i.e., $\{\tau(e_i)\}_1^n$ are in $W$, probably that's why they can not be treated as coefficients?
In any case, from either $\tau(v)=\sum \tau(e_i)\cdot e_i^*(v)\ \ $ or $\ \ \tau(v)=\sum e_i^*(v)\cdot \tau(e_i)$, it's not obvious to see that $\tau$ is a linear combination of a basis in $\hom(V,W)$, as the dimension of $\hom(V,W)$ is $m\times n$ while the sum only goes from $1$ to $n$. So my question is, is there a way to express $\tau$ as the linear combination of a basis in $\hom(V,W)$, similar to what we see for the linear functional $f\in \hom(V,\mathbb{R})$?
Thanks,
/bruin
Fix a basis $\mathcal{F}=\{f_1,f_2,\ldots,f_m\}$ of $W$. Then, for each $i\in\{1,2,\ldots,n\}$, $\tau(e_i)$ can be written as $\sum_{j=1}^ma_{ij}f_j$. Therefore\begin{align}\tau(v)&=\sum_{i=1}^ne_i^*(v)\tau(e_i)\\&=\sum_{i=1}^n\sum_{j=1}^ma_{ij}e_i^*(v)f_j.\end{align}You can deduce from this that the elements of $\hom(V,W)$ of the form $v\mapsto e_i^*(v)f_j$ form a basis of $\hom(V,W)$.