I'm trying to review and teach myself functional analysis. I've done it years ago and trying to get back into it, and I did this proof and was wondering if it is correct or if I'm missing something.
Let $X$ and $Y$ be finite dimensional linear spaces equipped with bases $\Phi = \{\phi_1,..,\phi_m\}$ and $\Psi = \{\psi_1,...,\psi_n\}$, respectively, and let $T \in L(X,Y) = B(X,Y)$.
When $x = \sum_{j = 1}^m x_j\phi_j,$ show that $Tx = \sum_{i = 1}^n y_i\psi_i$ where $\vec y = \textbf{T}\vec x$ and $\textbf{T} = [\psi_i'(T(\phi_j)]_{1\leq i \leq n, 1\leq j \leq m}$, and $\Psi' = \{\psi_1',...,\psi_n'\}$ is the basis for $Y'$ that is dual to $\Psi$. The matrix $\textbf{T}$ is the representation of $T \in L(X,Y)$ equipped with $\Psi$ or $\Phi$, respectively.
Proof
The dual space of $Y$, $Y' = B(Y,\mathbb{F})$ These are continuous bounded functions from $Y$ to $F$. The basis of $Y'$, $\Psi'$ is given by bounded functions which span the whole set of bounded functions from $Y'$. $T(\pi_j) \in Y$, and $\psi'(T(\phi_j))\in \mathbb{F}$. This is just an element in $\mathbb{F}$. So you could see it as a coefficient for a vector in any vector space.
$\textbf{T}$ is a matrix, given by: $\textbf{T} = [\psi_i'(T(\phi_j)]_{1\leq i \leq n, 1\leq j \leq m}$. A matrix this would look like:
$$ \textbf{T} = \begin{bmatrix} \psi_1 '(T\psi_1) & \psi_1'(T\psi_2) & ... & \psi_1'(T\psi_m) \\ \psi_2'(T\psi_1) & ... & ... & ... \\ ... & ... & ... & ... \\ \psi_n'(T\psi_1) & ... & ... & ... \end{bmatrix}. $$
So let's call the matrix entries $\textbf{T}_{ij} = \psi_i'(T(\phi_j)$. $$ Tx = T\sum_{j = 1}^m x_j\phi_j = \sum_{j = 1}^m x_jT(\phi_j). $$ Then, $T(\phi_j)$ is some vector in $Y$ given by $T(\phi_j) = \sum_{i = 1}^n \alpha_{ij} \psi_i$. Observe that $$ \alpha_{ij} = \psi_i'(\sum_{i = 1}^n \alpha_{kj} \psi_k) = \psi_i'(T(\phi_j)) = \mathbf T_{ij}. $$ Finally, if $\vec y = \mathbf T \vec x$, then $y_i = \sum_{j=1}^n \mathbf T_{ij} x_j$. From this, we get $$ Tx = T\sum_{j = 1}^m x_j\phi_j = \sum_{j = 1}^m x_jT(\phi_j) = \sum_{j = 1}^m x_j\sum_{i = 1}^n \mathbf T_{ij} \psi_i = \sum_{i=1}^n (\sum_{j=1}^m \mathbf T_{ij} x_j)\psi_i = \sum_{i=1}^n y_i\psi_i $$
Your proof is correct. However, it could be presented more clearly and concisely. Here's an edit to demonstrate what I mean.
Let $\mathbf T_{ij} = \psi_i'(\phi_j)$ denote the entries of $\mathbf T$. For each $j = 1,\dots,m$, we observe that (by the definition of a dual basis) $$ T(\phi_j) = \sum_{i=1}^n \psi_i'(T(\phi_j))\psi_i = \sum_{i=1}^n \mathbf T_{ij}\psi_i. $$ If $\vec y = \mathbf T \vec x$, then $y_i = \sum_{j=1}^n \mathbf T_{ij} x_j$. With this in mind, we observe that $$ Tx = T\sum_{j = 1}^m x_j\phi_j = \sum_{j = 1}^m x_jT(\phi_j) = \sum_{j = 1}^m x_j\sum_{i = 1}^n \mathbf T_{ij} \psi_i = \sum_{i=1}^n \left(\sum_{j=1}^m \mathbf T_{ij} x_j\right)\psi_i = \sum_{i=1}^n y_i\psi_i, $$ as desired.