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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 $(1,2,3)$

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4 Answers 4

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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.

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  • $\begingroup$ Exactly what I needed. I was confused because I could express $$x_2$$ as $\frac{y_3}{3}$ and as $\frac{y_1-y_2}{2}$. But now I see that this is a constrain. $\endgroup$ Dec 5, 2016 at 8:55
  • $\begingroup$ @RaducuMihai Great. You have got the point. $+1$ for your question too! $\endgroup$ Dec 5, 2016 at 9:07
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The image of a linear transformation are the vector in $\mathbb{R}^3$ such that $(a,b,c)=(x_1+x_2,x_1-x_2,3x_2)=(x_1,x_1,0)+(x_2,-x_2,3x_2)=x_1(1,1,0)+x_2(1,-1,3)$, i.e., $Im(T)=\left\langle \{(1,1,0),(1,-1,3)\} \right\rangle$, because clearly the vectors are linearly independent.

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The vector space $\mbox{Im}(T)$ has dimension $2$ and it is generated by the linearly independent vectors $(1,1,0)$ and $(1,-1,3)$. Hence a vector $(y_1,y_2,y_3)\in \mbox{Im}(T)$ iff $3(y_1-y_2)=2y_3$.

Since $3(1-2)\neq 2\cdot 3$, the vector $(1,2,3)$ does not belong to $\mbox{Im}(T)$.

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It looks like your solution for the $x$'s is incorrect.

Observe that by subtracting the first two equations and solving for $x_2$, we get $x_2 = \frac{y_1 - y_2}{2}$. The third equation tells us $x_2 = \frac{y_3}{3}$. Set both of these expressions we just found for $x_2$ equal to each other, to obtain $\frac{y_1 - y_2}{2} = \frac{y_3}{3}$. This gives you the relationship that $y_1$, $y_2$, and $y_3$ must have relative to each other.

If we solve for $y_3$ (the easiest of all the $y$'s to solve for), we get $y_3 = \frac{3}{2}(y_1 - y_2)$. So your vector $y$ becomes

$y = \left(\begin{matrix}y_1\\y_2\\y_3\end{matrix}\right) = \left(\begin{matrix}y_1\\y_2\\\frac{3}{2}y_1 - \frac{3}{2}y_2\end{matrix}\right) = \frac{y_1}{2}\left(\begin{matrix}2\\0\\3\end{matrix}\right) + \frac{y_2}{2}\left(\begin{matrix}0\\2\\-3\end{matrix}\right)$

But $y_1$ and $y_2$ can be any values we'd like, so what we really have is an arbitrary linear combination of the two vectors $(2, 0, 3)$ and $(0, 2, -3)$. Hence, we can say that $\text{Im}(T) = \text{span}\{(2, 0, 3), (0, 2, -3)\}$, i.e. every vector "produced" by $T$ can be expressed as some weighted sum of these two vectors.

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