Is the Range of This Linear Transformation the Entire R2?

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SUMMARY

The linear transformation T: R^3 → R^2 defined by T(v_1, v_2, v_3) = (v_3 - v_1, v_3 - v_2) has a kernel that is one-dimensional, represented by the basis vector s(1, 1, 1). The dimension of the range of T is confirmed to be two, as the dimension of the kernel plus the dimension of the range equals the dimension of the domain (3). The range is a subspace of R^2, which inherently has only one two-dimensional subspace, indicating that the range of T is indeed the entire R^2.

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Homework Statement



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*}&amp;v_3 - v_1 = 0\\<br /> &amp;v_3 - v_2 = 0\end{align*}

Thus, \begin{align*}&amp;v_3 = v_1 \\<br /> &amp;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?
 
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Everything seems fine. So where's your difficulty?
 
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!
 

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