A linear functional on the space of transformations is basis independent I've been working on this problem for a bit and am not sure how to proceed:  let $V$ be an $n$ dimensional $\mathbb{ R } $-vector space, and denote by $\mathcal{L}(V)$ the space of linear operators $V \to V$.  Denote by $V^*$ the dual space of $V$.  For every basis $B = \{ v_{1}, \ldots , v_{n}\}$ we construct the dual basis $B^* = \{ v^*_{1}, \ldots , v^*_{n} \}$ of $V^*$ such that 
$$
 v_i ^* (v_j) = \delta_{i,j} 
$$
where $\delta_{i,j} = 1$ if $i = j$ and $0$ otherwise.  For $T \in \mathcal{L} (V)$, define
$$
 \varphi (T) = \sum_{i=1}^{n} v_i ^* (T(v_i))
$$
The problem asks to show that $\varphi : \mathcal{L}(V) \to \mathbb{ R } $ is a linear functional, which I believe I've done (it essentially follows from linearity of the dual basis vectors $v_i ^*$).  The problem ask further for us to show that the value $\varphi (T)$ is independent of the choice of basis $B$ for $V$.  This is the part which is giving my difficulty.  I have expressions for $\varphi(T)$ with respect to two different bases for $V$ but don't see why they must be the same.  Furthermore the hypothesis that $V$ is a real vector space is throwing me off-is this  important?  Perhaps this is where the key lies...I would greatly appreciate any help!   
 A: Let $\{w_j\}_{j=1}^n$ is another basis of $V$ and let $v_i=\sum\limits_{j=1}^{n}{\alpha_{ij}w_j},\,\, i=1,..,n$  is the representation of the first basis in terms of the second.
$\varphi(T)=\sum\limits_{i=1}^{n}{v_i^*(T(v_i))}\\
=\sum\limits_{i=1}^{n}{v_i^*(T(\sum\limits_{j=1}^{n}{\alpha_{ij}w_j}))}\\
=\sum\limits_{i=1}^{n}{v_i^*\left(\sum\limits_{j=1}^{n}{\alpha_{ij}T(w_j)}\right)}\\
=\sum\limits_{i=1}^{n}{\sum\limits_{j=1}^{n}{\alpha_{ij}v_i^*(T(w_j))}}\\
=\sum\limits_{j=1}^{n}{\sum\limits_{i=1}^{n}{\alpha_{ij}v_i^*(T(w_j))}}\\
=\sum\limits_{j=1}^{n}{w_j^*(T(w_j))}$
The last step is because $\sum\limits_{i=1}^{n}{\alpha_{ij}v_i^*(a)}=w_j^*(a)$ for any vector $a$ in $V$. To show this it is only needed to show that it holds for each vector from some basis (because of linearity). For example, show it for $\{v_i\}_{i=1}^n$:
$\sum\limits_{i=1}^{n}{\alpha_{ij}v_i^*(v_k)}=\sum\limits_{i=1}^{n}{\alpha_{ij}\delta_{ik}}=\alpha_{kj}$
$w_j^*(v_k)=w_j^*\left(\sum\limits_{l=1}^{n}{\alpha_{kl}w_l}\right)=\sum\limits_{l=1}^{n}{\alpha_{kl}w_j^*(w_l)}=\sum\limits_{l=1}^{n}{\alpha_{kl}\delta_{jl}}=\alpha_{kj}$
