I'm working on the following problem:
Define a linear transformation $f: \mathbb{R}^3 \to \mathbb{R}^3$ such that:
- $\{x \in \mathbb{R}^3: f(x) = (1, 1, 1)\} = \{x \in \mathbb{R}^3: x_1 + x_2 + x_3 = 1, x_2 - x_3 = 2\}$
- $Im(f) = \langle(1, 2, 3), (0, 1, 2)\rangle$.
Solving the equations for the preimage of $(1,1,1)$, I get that $f^{-1}(1,1,1) = \langle(-2, 1, 0)\rangle + (-1, 2 0)$. In case the notation isn't clear, this represents the line parallel to $(-2,1,0)$ that goes through $(-1,2,0)$. The equation $f(x) = (1,1,1)$ is a non homogeneous linear system. Therefore, $\langle(-2,1,0)\rangle$ must be the solution to $f(x) = 0$, and $f(-1,2,0)$ must be $(1,1,1)$. This tells us right away that the kernel of $f$ is $\langle(-2,1,0)\rangle$. We can begin writing the values of $f$ for some basis of $\mathbb{R}^3$:
$\begin{align} f(-2,1,0) &= 0 \\ f(-1,2,0) &= (1,1,1) \end{align} $
Now, I would think that the only restriction I have for the third vector is that its image should complete a basis for $Im(f)$. We can use, for example, $(1,3,5)$, because $\langle(1,1,1),(1,3,5)\rangle = \langle(1,2,3),(0,1,2)\rangle = Im(f)$. As for the third element of the basis of $\mathbb{R}^3$, I don't see any particular restriction for it, so I'll just use $(0,0,1)$. We can define $f$ like this, then:
$\begin{align} f(-2,1,0) &= 0 \\ f(-1,2,0) &= (1,1,1) \\ f(0,0,1) &= (1,3,5) \end{align}$
This looks all right. But if I change to the canonical basis, write the matrix of $f$ and try to solve the system $f(x) = (1,1,1)$, I get something different than what it should be. I even used WolframAlpha to make sure I hadn't messed up when solving the linear systems. What am I doing wrong here?