let $V,W$ be vector spaces with a finite dimension. Let $v_1,...,v_n$ be a basis of $V$ and $w_1,...,w_m$ a basis of $W$ respectively. Then $(v_1,0),...,(v_n,0),(0,w_1),...,(0,w_m)$ forms a basis of $V\oplus W$. Lastly, we have $i: V\to V\oplus W, v\mapsto (v,0)$.
I have to find a transformation matrix for the aforementioned basis, any ideas?
Consider the $k$-vector spaces $V$ and $W$ with bases $\mathscr B_V = \{v_1, \dots, v_n \}$ and $\mathscr B_W = \{w_1, \dots, w_m \}.$ Like you mentioned, the vectors $u_i$ whose first $n$ coordinates are identical to $v_i$ and whose last $m$ coordinates are $0$ for $1 \leq i \leq n$ and whose first $n$ coordinates are $0$ and whose last $m$ coordinates are identical to $w_i$ for $n + 1 \leq i \leq n + m$ form a basis for $V \oplus W.$ Explicitly, we have that $a_1 u_1 + \cdots + a_{n + m} u_{n + m} = 0_{n + m}$ if and only if $a_1 v_1 + \cdots a_n v_n = 0_n$ and $a_{n + 1} w_{n + 1} + \cdots + a_{n + m} w_{n + m} = 0_m$ if and only if $a_1 = \cdots = a_{n + m} = 0.$
Consequently, a linear operator $T : V \oplus W \to V \oplus W$ is uniquely determined by the images $T(u_1), \dots, T(u_{n + m}).$ Considering that each of these images is a vector of $V \oplus W,$ we have that $$\begin{align*} T(u_1) &= a_{1, 1} u_1 + a_{2, 1} u_2 + \cdots + a_{n + m, 1} u_{n + m} \\ \\ T(u_2) &= a_{1, 2} u_1 + a_{2, 2} u_2 + \cdots + a_{n + m, 2} u_{n + m} \\ &\phantom{\,\,} \vdots \\ T(u_{n + m}) &= a_{1, n + m} u_1 + a_{2, n + m} u_2 + \cdots + a_{n + m, n + m} u_{n + m}\end{align*}$$ for some constants $a_{ij}$ in $k.$ We can therefore write the matrix of $T$ with respect to the basis $\mathscr B = \{u_1, \dots, u_{n + m} \}$ as the $(n + m) \times (n + m)$ matrix whose $(i, j)$th element is $a_{ij}$ for $1 \leq i, j \leq n + m.$