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
$B=\{(0,0)(-1/4,-1/2)\} = \{(-1/4,-1/2\}$
Best Answer
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 $v_2$.