# How to find the matrix for a linear operator in respect to the basis vectors?

The question is:

Suppose $V = U_1\oplus U_2$. Define the linear operator $T$ on $V$ as follows: for every $v$ find $u_1\in U_1$ and $u_2\in U_2$ such that $v = u_1+u_2$, then $T(v)=u_1$.

Also, the basis are given for $U_1$ and $U_2$.

$B=\{b_1,\cdots,b_r\}$ and $C=\{c_1,\cdots,c_{n-r}\}$.

How would I prove that $B \bigcup C$ is a basis for $V$? And how would I write the matrix of $T$ in respect to $B \bigcup C$? The professor wrote it as $[T]_{B \bigcup C}$.

Is the matrix for $T$ = $\begin{bmatrix} 1 & 0 \end{bmatrix}$?

Because $T(v) = u_1\rightarrow T(u_1+u_2) = u_1 \rightarrow \begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} = 1(u_1) + 0(u_2) = u_1$? How would I represent this transformation matrix in respect to $B$ and $C$?

$\alpha_1b_1+ ... + \alpha_rb_r+\alpha_{r+1}c_1+...+\alpha_nc_{n-r} = 0$ then all the scalars are zero. (Hint: apply T to both sides and use the fact that B and C are already bases). Using the argument that you have n linearly independent vectors and both B and C are a subset of V, then you have a basis with equal dimension of the vector space, therefore it is a basis for the vector space.
as for $[T]_{B \bigcup C}$ matrix representation has columns of T applied to every basis vector, written in the basis of ${B \bigcup C}$. Since $T(v)$ outputs the B component of every vector in V, the columns of T will be zero when applied to basis vectors in C, and it will be diagonal when applied to vectors in basis B. Note that the basis representation cannot be $\begin{bmatrix} 1 & 0 \end{bmatrix}$. Think about it in terms of dimensions to figure out why (clearly, T must be n dimensional)