Given linear transformation T, prove $N(T)=\ker(T)$ and $R(T)=\mathrm{range}(T)$

Question

$ \newcommand{\Ker}{\mathrm{Ker}} \newcommand{\Range}{\mathrm{Range}} $Let us define a linear transformation $T:\mathbb{R}^3 \rightarrow \mathbb{R}^2,$ by $T(a_1, a_2, a_3)=(a_1 - a_2, 2a_3).$ I would like to prove that $$N(T)=\Ker(T)=\{(a,a,0):a\in \mathbb{R} \}$$ and $$R(T)=\Range(T)=\mathbb{R}^2.$$

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Let me start with what we know.

We say that a function $T:V \rightarrow W$ is a linear transformation provided that $$T(u+v)=T(u)+T(v)$$ and $$T(av)=aT(v), \forall v \in V, a\in F.$$ We have that the null-space or kernel of a linear transformation is defined as $$N(T)=\{x \in V : T(x)=0 \}$$ where $V$ in this case is $\mathbb{R}^3$ and the range or image of linear transformation is defined as $$R(T)=\{T(x) \in W: x\in V \},$$ where $W$ here represents $\mathbb{R}^2$.

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It seems as though the answer is quite obvious, but I am struggling creating a formal proof around this. Do we need to first show that this is indeed a linear transformation? Next, how do we concretely show that these are the proper cases. Thank you for your help!

Answer 1

Hint:

• Suppose $(a_1,a_2,a_3)\in N(T)$, then $$T(a_1,a_2,a_3)=(a_1-a_2,2a_3)=(0,0)$$ What can you say about $(a_1,a_2,a_3)$ for this to be true?

• To prove that $R(T)=\Bbb R^2$, can you choose $a_1,a_2,a_3$ such that $(a_1-a_2,2a_3)$ hits everything in $\Bbb R^2$? In particular, if I choose a point $(x,y)\in\Bbb R^2$, can you find $(a_1,a_2,a_3)$ that maps to $(x,y)$?

Answer 2

First, \begin{align} (a_1,a_2,a_3)\in N(T)&\iff T(a_1,a_2,a_3)=0\\ &\iff (a_1-a_2,2a_3)=(0,0)\\ &\iff a_1=a_2\mbox{ and }a_3=0. \end{align}

Next, let $(x,y)\in \mathbb{R}^2$. Then $(x,y)=T(x,0,y/2)$. So $(x,y)\in R(T)$. As $(x,y)$ was arbitrary, $R(T)=\mathbb{R}^2$.