- #1

- 2

- 0

## Homework Statement

T:[tex]{R^3 \rightarrow {R^2}[/tex] given by [tex]T(v_1,v_2,v_3) = (v_3 -v_1, v_3 - v_2)[/tex]

If linear, specify the range of T and kernel T

**The attempt at a solution**

Okay, I went ahead and tried to find the kernel of T like here:

[tex]\begin{align*}&v_3 - v_1 = 0\\

&v_3 - v_2 = 0\end{align*}[/tex]

Thus, [tex]\begin{align*}&v_3 = v_1 \\

&v_3 = v_2\end{align*}[/tex]

So choosing v3 as s gives the 1-D basis of W= s(1, 1, 1) **a column vector**

But I'm not entirely sure how to get the range. IF I did the kernel correctly, then that means the dimension of the range will be 2 as 2+1 = 3 (the dimension of the domain). But when I try to do the range, I get a 3-dimensional basis where v1,v2,and v3 are their own LI vectors as so:

(y1,y2) = s(1,1) + t(-1,0) + r(0,-1)

Any help?