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The question states "Find the range and the kernel (those are new words for the column space and nullspace) of $T$.

Part a is:

$$T(v_1, v_2) = (v_2, v_1)$$

Is the kernel

$$\begin{bmatrix} 0 \\ 0 \end{bmatrix}$$

because the only way to get $(v_2, v_1) = 0$ is for both $v_1$ and $v_2$ to be $0$, and the range

$$\begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$

due to the fact that the columns space becomes

$$v_2 \begin{bmatrix} 1 \\ 0 \end{bmatrix}+ v_1 \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$

and in a related note, if the original question becomes

$$T(v_1,v_2,v_3) = (v_1,v_2)$$

does $v_3$ in both the kernel and range become irrelevant due to the fact $v_3$ seems to become a free variable? In othe words, the kernel is $(0, 0, 1)$ and the range is $(1, 0, 0)$ and $(0, 1, 0)$.

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up vote 1 down vote accepted

I'm not sure if I understand you right, but if you mean that the range is the entire space $K^2$ (with $K$ the base field), and the kernel is $\{0\}$ then yes, I believe you are correct.

I'm not sure what you mean by that $v_3$ becomes free variable, but you're correct in stating that it spans the kernel (which is $\{(0,0,x)\vert x\in K\}$) and that the range is still entire $K^2$ (of course, it can't be anything more!).

Remember, however, that the kernel and the range are spaces, sets of vectors. Those you've written down are just the spanning ones, not the entire spaces, so you should write $\operatorname{rng} T= \operatorname{lin}\left(\begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix}\right)$ etc. (or however else you denote the linear span).

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Yes, that is what my intention was. Should I write span({1,0}, {0,1}) to make that clearer for future reference and the graders? EDIT: Ah, I see. I do not believe we have gone over that notation in class yet so unless span implies something different, I believe it will suffice. – Goyatuzo Oct 12 '12 at 0:59
@Goyatuzo: yes, you should not omit the span, to make your statements actually formally correct, if for no other reason. – tomasz Oct 12 '12 at 1:01

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