I'm being asked a basis $B$ such that $|f|_B = \begin{bmatrix} 0&0\\ 0&1\\ \end{bmatrix} $
$ f: R^2 \rightarrow R^2 $ is linear transformation that is a projection such that
$ f([x_1 x_2]^T) = [2x_1+x_2 \ \ \ -2x_1-x_2]^T $
Then, I define as the basis I will work in...
$ N = \{(1,0)(0,1)\} $
Then, the linear transformation does...
$ f([1 0]^T) = [2 \ \ -2]_B $
$ f([0 1]^T) = [1 \ \ -1]_B $
The matrix $A$ associated to the linear transformation Ax=b is then:
$A= \begin{bmatrix} 2 & 1\\ -2 & -1\\ \end{bmatrix} $
If I'm understanding right, I'm being asked a pair of vectors $v_1 , v_2 \in B$ such that
$ Av_1 = [0 0]^T \\ Av_2 = [0 1]^T $
But when trying to solve it this way, I end up with a couple of equations that I'm not understading well, also, I'm not using the additional information that this linear transformation is a projection.
Edit because of hint
So I can choose whatever solution I want to...
Let $v_1 = (a,b)$ and $v_2 = (c,d)$
$ \begin{bmatrix} 2&1\\ -2&-1\\ \end{bmatrix} $ $ \begin{bmatrix} a\\ b\\ \end{bmatrix} $= $ \begin{bmatrix} 0\\ 0\\ \end{bmatrix} $ I can say $v_1 = (0,0)$
$ \begin{bmatrix} 2&1\\ -2&-1\\ \end{bmatrix} $ $ \begin{bmatrix} c\\ d\\ \end{bmatrix} $= $ \begin{bmatrix} 0\\ 1\\ \end{bmatrix} $ I can say $v_2 = (-1/4,-1/2)$
Therefore $B=\{(0,0)(-1/4,-1/2)\} = \{(-1/4,-1/2\}$
Hint You are being asked to find two vectors $v_1$ and $v_2$ such that $f(v_1)=0$ and $f(v_2)=v_2$. Let $v_1=(x_1,x_2)$ and solve $f(x_1,x_2)=(0,0)$ and choose one solution (there are infinitely many solutions). The same for $v_2$.