Linear transformation one-to-one

jonroberts74
Messages
189
Reaction score
0

Homework Statement


let ##T:\mathbb{R^3} \rightarrow \mathbb{R^3}## where ##T<x,y,z>=<x-2z,y+z,x+2y>##

Is T one-to-one and is the range of T ##\mathbb{R^3}##?

The Attempt at a Solution



I took the standard matrix A ##\left[\begin{array}{cc}1&0&-2\\0&1&1\\1&2&0\end{array}\right]##

det(A)=0 so by equivalence T is not one-to-one and by equivlence again the range is not ##\mathbb{R^3}##
 
Physics news on Phys.org
What is your question?
 
is that correct?
 
I suspect you already know the answer is yes.
 
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