I need to find the $Im(T)$, where $T:\mathbb{R^2} \rightarrow \mathbb{R^3}; T(x_1,x_2) = (x_1+x_2,x_1-x_2,3x_2)$.
I know that $$Im(T) = \{y \in \mathbb{R^3}| \exists x \in \mathbb{R^2} \ T(x) = y\}$$
So I need to solve the system: $$\left\{\begin{matrix}
x_1+x_2=y_1\\
x_1-x_2=y_2\\
3x_2=y_3
\end{matrix}\right.$$
And I get the solution $x_1 = y_1 – \frac{y_3}{3}\ and\ x_2 =\frac{y_3}{3}$.
My problem is that if I try to find $x_1$ and $x_2$ for the vector $y = (1,2,3)$ I get, for example; $x_1 = 0$ and $x_2 = 1$. But $T(0,1)$ is not $(1,2,3)$
Best Answer
The image of $T$ is basically all elements of the form $Tx$, $x$ a two dimensional real vector.
So let $y = (y_1,y_2,y_3)$. If $y = T(x_1,x_2)$, then the following must be satisfied:
1) $x_1 +x_2 = y_1$
2) $x_1 - x_2 = y_2$
3) $3x_2 = y_3$
Solve these equations, you get $x_1 = \frac{y_1+y_2}{2}$, $x_2 = \frac{y_3}{3} = \frac{y_1-y_2}{2}$
Now, you have to ask the following question: Can $y_1,y_2,y_3$ be chosen independently? What are the constraints binding them?
You can see that there is only one constraint: $\frac{y_3}{3} = \frac{y_1-y_2}{2}$. The other component $\frac{y_1+y_2}{2}$ can move around freely, so we need not worry about that.
Hence, the range is all $(y_1,y_2,y_3)$ such that $2y_3 = 3(y_1-y_2)$. You can check that this is a two-dimensional subspace, spanned by the vectors $(1,1,0)$ and $(3,1,3)$, for example.