Linear Transformation R2->R3 with 'zero' vector

A440
Messages
1
Reaction score
0

Homework Statement


Is T(X,Y)->(X,Y,1) a linear transformation? where X and Y are defined R2 column vectors.

Homework Equations


Attempt to prove T(cX+Y)=cT(X)+T(Y)
Consider T(cx1+y1,cx2+y2)->(cx1+y1,cx2+y2,1)

The Attempt at a Solution


RS=cT(x1,y1)+T(x2,y2)->c(x1,y1,1)+(x2,y2,1)
=(cx1+y1,cx2+y2,1)+(0,0,c)
=(cx1+y1,cx2+y2,1)+c(0,0,1)
The 'zero' vector: T(0,0)->(0,0,1)
therefore T(cx1+y1,cx2+y2)->(cx1+y1,cx2+y2,1)
and T(X,Y)->(X,Y,1) is a linear transformation.
 
Last edited:
Physics news on Phys.org
zero vectors are only of the form (0,0) and (0,0,0).
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top