# Determine matrix A with respect to standard basis $f:U->R^2$

I am having trouble understanding my problem and what to calculate

I have been given the subspace $$U=\{x= \begin{vmatrix} x_1\\ x_2\\ x_3\\ \end{vmatrix} \in F^3 | x_1 + x_2 + x_3 = 0\} \subset F^3$$

and the linear transformation $$f: U \rightarrow F^2$$ $$f\begin{vmatrix} x_1\\ x_2\\ x_3\\ \end{vmatrix} =\begin{vmatrix} x_1\\ x_2+x_3\\ \end{vmatrix}$$

The question is to determine a matrix A that represent $$f: U \rightarrow F^2$$ with respect to the basis for U and standard basis $$(e_1,e_2)$$ for $$F^2$$

My attempt: I have calculated the basis $$B=\{\begin{vmatrix} 1\\ 0\\ -1\\ \end{vmatrix},\begin{vmatrix} 0\\ 1\\ -1\\ \end{vmatrix}\}$$

And my matrix A calculated from the linear transformation

$$\begin{vmatrix} 1&0&0\\ 0&1&1\\ \end{vmatrix}$$

I'm aware the standard basis are $$e_1=\begin{vmatrix} 1\\ 0\\ \end{vmatrix} , e_2=\begin{vmatrix} 0\\ 1\\ \end{vmatrix}$$

I'm not sure about the next step. Do I calculate: $$f\begin{vmatrix} 1\\ 0\\ -1\\ \end{vmatrix} =\begin{vmatrix} 1\\ -1\\ \end{vmatrix}$$

and do the same thing for the other basis or am I suppose to find a matrix A that's going to give me $$f\begin{vmatrix} 1\\ 0\\ -1\\ \end{vmatrix} =\begin{vmatrix} 1\\ 0\\ \end{vmatrix}$$

and

$$f\begin{vmatrix} 0\\ 1\\ -1\\ \end{vmatrix} =\begin{vmatrix} 0\\ 1\\ \end{vmatrix}$$

I just don't understand the question. How am I suppose to find a matrix A with respect to the basis of U and the standard basis for $$F^2$$?

The image of the first vector in the basis is $$\begin{bmatrix} 1 \\ 0+(-1) \end{bmatrix}$$ and the image of the second vector is $$\begin{bmatrix} 0 \\ 1+(-1) \end{bmatrix}$$ so the matrix is $$\begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix}$$ Note that the map $$f$$ can be more easily described as $$f\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} x_1 \\ -x_1 \end{bmatrix}$$
• @mahma The map is $f\colon U\to F^2$ and the domain has dimension $2$, as well as the codomain; so the representing matrix must be $2\times2$. – egreg Jan 3 at 0:47