Is the Range of This Linear Transformation the Entire R2?

Click For Summary
The linear transformation T maps R^3 to R^2 and is defined as T(v1, v2, v3) = (v3 - v1, v3 - v2). The kernel of T is determined to be one-dimensional, represented by the vector s(1, 1, 1). The dimension of the range is inferred to be two, as the sum of the dimensions of the kernel and range equals the dimension of the domain. However, there is confusion regarding the basis for the range, as attempts yield a three-dimensional basis, which contradicts the earlier conclusion. Ultimately, it is clarified that the range of T is indeed the entire R^2, as it can accommodate all possible outputs in that space.
oomba
Messages
2
Reaction score
0

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?
 
Physics news on Phys.org
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!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K