Finding the Matrix of a Linear Transformation

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Homework Statement



Assume that T defines a linear transformation and use the given information to find the matrix of T
T: R4-->R2
such that T(1,0,0,0)=(3,-2), T(1,1,0,0)=(5,1), T(1,1,1,0)=(-1,0), and T(1,1,1,1)=(2,2)

Homework Equations


The Attempt at a Solution



I think I need to use/find the standard basis for R4, but not really sure.

Thanks for the help.
 
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The standard basis for R^4 is (1,0,0,0), (0,1,0,0), (0,0,1,0), and (0,0,0,1). The idea of this problem is that once you know where the basis elements go, you can uniquely determine what the matrix is (using methods described in your textbook!)

So you need to find T(1,0,0,0), which you already know, along with T(0,1,0,0), T(0,0,1,0), and T(0,0,0,1). How would you find these? (Hint: Use the definition of linearity.)
 
Do I set them as linear combinations of each other? [a b c d]=c1v1+c2v2+c3v3+c4v4 ?
 
OIWQJFEOIWJQEF just figured it out.

for example, in order to find (0,1,0,0), we can say that's

(0,1,0,0) = a(1,1,0,0)-b(1,0,0,0) if we set a=b=1

T(0,1,0,0)=T(1,1,0,0)-T(1,0,0,0)
T(0,1,0,0)=(5,1)-(3,-2)
T(0,1,0,0)=(2,3) which is indeed the answer provided. I just have to do this for the rest of the vectors.thanks for the help :smile:
 
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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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