T is finite rank operator represent [closed]

Let $\{b^j\}_{j=1}^n$ be the standard dual basis of $T(E)^*$ with respect to $\{b_j\}_{j=1}^n$. Now, for every $j\in\{1,\ldots,n\}$, define $\phi_j:E\rightarrow \mathbb{K}$ by $$ \phi_j=b^j\circ T.$$ Writing $$ T(x)=\displaystyle\sum_{i=1}^n \alpha_ib_i,$$ we have that \begin{align*}\phi_j(x)&=b^j\left(T(x)\right)\\ &=b^j\left(\displaystyle\sum_{i=1}^n\alpha_ib_i\right) \\ &=\displaystyle\sum_{i=1}^n\alpha_ib^j(b_i)\\ &=\displaystyle\sum_{i=1}^n\alpha_i\delta^i_j\\ &=\alpha_j. \end{align*} Therefore, indeed $$ T(x)=\displaystyle\sum_{i=1}^n\phi_i(x)b_i.$$ Finally, since each $b^j$ is a linear map defined over finite dimensional spaces (therefore bounded), and since $T$ is assumed to be a bounded linear map, it is immediate that each $\phi_j$ is linear and bounded (composition of linear bounded maps).