Finding the Matrix Relative to Basis B

  • Thread starter Thread starter mollyduncanst
  • Start date Start date
  • Tags Tags
    Bases Matrix
mollyduncanst
Messages
1
Reaction score
0

Homework Statement


Let T be the linear transformation T=[1, 3, 2, 6] (the matrix has 1 and 3 on the top row and 2 and 6 on the bottom row) relative to the standard basis

Find the matrix relative to the basis B= {(1, 2), (-3, 1)}

Homework Equations


The Attempt at a Solution


So what I did was apply the basis vector 1 to the matrix and then apply the basis vector 2 to the matrix again and got (7, 14) and (0, 0) but I don't know where to go from here, what do I do with these 2 vectors?
 
Physics news on Phys.org
I am not sure, but when saying T=[1,3; 2,6] is relative to E, it means that
T*e1^t=(1,2)
T*e2^t=(3,6)
Am I correct?
If I have remembered correctly, then we want to find T' such that
T'*(1,2)^t=(1,2)
T'*(-3,1)^t=(3,6)
or in other words
[a,b;c,d]*(1,2)^t=(1,2) => (a+2b,c+2d)=(1,2)
[a,b;c,d]*(-3,1)^t=(3,6) => (-3a+b,-3c+d)=(3,6)
which is actually:
[1] a+2b=1
[2] -3a+b=3
[3] c+2d=2
[4] -3c+d=6
 
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