# Linear transformations

1. May 10, 2008

### oomba

1. The problem statement, all variables and given/known data

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

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:
\begin{align*}&v_3 - v_1 = 0\\ &v_3 - v_2 = 0\end{align*}

Thus, \begin{align*}&v_3 = v_1 \\ &v_3 = v_2\end{align*}

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?

2. May 11, 2008

### foxjwill

Everything seems fine. So where's your difficulty?

3. May 11, 2008

### HallsofIvy

Staff Emeritus
You have correctly deduced that the range must have dimension 2 and you know that the range is a subspace of R2.

How many subspaces of dimension 2 do you think R2 has!