Orthogonal projection onto the line $y = 2x$ gives a linear transformation $T: R2 → R2$ such that
$$T(1,2) = (1,2)$$ and $$T(−2,1) = (0,0)$$
Then the matrix of T with respect to the ordered basis
$$B = \{(1, 2), (−2, 1)\}$$ and the standard basis $$S = \{(1, 0), (0, 1)\}$$ is?
Attempt to solve for ordered? (or is it for standard?): $$\begin{bmatrix} 1 & -2\\ 2 & 1 \end{bmatrix}*T=\begin{bmatrix} 1 & 0\\ 2 & 0 \end{bmatrix}$$ solving for T:
$$T=\begin{bmatrix} 1 & -2\\ 2 & 1 \end{bmatrix} ^{-1} *\begin{bmatrix} 1 & 0\\ 2 & 0 \end{bmatrix}$$
$$T=\begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix}$$ when I check it geometrically (for projection), it always gives me the zeros for the second coordinate, however is it supposed to be perpendicular projection of the point on the line y=2x.
Can you kindly tell me what's going wrong with it? Or my solution is completely incorrect? And what about getting T in the standard canonical basis? is it just the matrix T multiplied by Transformation matrix of B base to S base: $$P_{BS}$$ i.e. $$T*P_{BS}$$