Find the image of this linear transformation: $$f(x_1,x_2,x_3,x_4)=(x_1+x_2-x_3, x_1-x_2-x_3, x_1+x_2+x_3,x_2-x_3)$$
My very first thought was just to do the simple algebraic move: $$(x_1+x_2-x_3, x_1-x_2-x_3, x_1+x_2+x_3,x_2-x_3)=x_1(1,1,1,0)+x_2(1,-1,1,1)+x_3(-1,1,1,0)$$ So then we have that $\text{Im} f=\text{span}\{(1,1,1,0),(1,-1,1,1), (-1,1,1,0)\}$. But the book tries to solve the system formed by an arbitrary variable $y=(y_1, y_2, y_3, y_4)$ and does Gaussian simplification of the matrix equation $y=Fx$, being $F$ the matrix that represents the transformation, and they get to another whole different image, is that the way to go? Or mine is as well correct but there can be different images for the same linear transformation?
Best Answer
Your solution is correct.
The solution can have different ways to present, and they are all equivalent. Let's see a simple example, suppose $f(x_1, x_2)=(x_1, x_2)$, then
$$\text{Im}f=\text{span}\{(1,0), (0,1)\}=\mathbb R^2$$
but you can also write it as
$$\text{Im}f=\text{span}\{(1,1), (0,1)\}=\text{span}\{(e,\sqrt2), (-1,\pi)\}=\dots$$
Namely, as long as the solution spans $\mathbb R^2$, they are all correct.