Linear Algebra Find the Standard Matrix of T

x.x586
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Homework Statement



Let T be a linear transformation from R3 to R3. Suppose T transforms (1,1,0) ,(1,0,1) and (0,1,1) to (1,1,1) (0,1,3) and (3,4,0) respectively.

Find the standard matrix of T and determine whether T is one to one and if T is onto
 
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welcome to pf!

hi x.x586! welcome to pf! :wink:

show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
I know T(x) =Ax=[T(e1) ,T(e2,) T(e3)]

I thought A would just be the matrix with columns (1,1,1) (0,1,3) and (3,4,0), but then I realized that
(1,1,0) ,(1,0,1) and (0,1,1) are not the standard basis vectors for R3My book doesn't give any examples where we don't start with the standard basis vectors

Should I have started by taking a 3x3 matrix entries [x1,x2,x3;x4,x5,x6,x7,x8,x9] and multiply that by a 3x3 matrix with entries [1,1,0;1,0,1;0,1,1] and set that equal to a matrix with entries [1,0,3;1,1,4;1,3,0] and then got a system of equations from there by multiplying the left side out. And then set up an augmented matrix and used row reduction to find corresponding entries for A?
 
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hi x.x586! :smile:

(i'm not guaranteeing this is the quickest way, but …)

if you can express eg (1,0,0) as a(1,1,0) + b(1,0,1) + c(0,1,1), then you'll have the form you're familiar with :wink:
 
Thank you.
 
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...
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