Determinant of linear transformation

riordo
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Homework Statement



symmetric 2 × 2 matrices to V.Find the determinant of the linear transformation T(M)=[1,2,2,3]M+[1,2,2,3] from the space V of symmetric 2 × 2 matrices to V.




Homework Equations





The Attempt at a Solution


hi this is my first post so if I break a rule please let me know so I can correct the issue. also is there some way to present the problems the way they are given in the text.

in this problem would you add the two matrices together and then multiple by a single M? for example [1,2,2,3]+[1,2,2,3]=[2,4,4,6]M=(2M*6M)-(4M*4M)=-4?
 
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The obvious way to do it is to write T as a 4x4 matrix using a basis of the 2x2 matrices, but there may be a more clever way. You could also write it as a 3x3 using the fact the matrices are symmetric. Welcome to the forum, riordo. You originally wrote [1,2,2,3]M+M[1,2,2,3], right? And you do mean [1,2,2,3]=[[1,2],[2,3]] as a symmetric 2x2 matrix, yes?
 
yes..thank you.
 
what would the 4x4 matrix look like? if you add a 2x2 matrix to another 2x2 matrix the solution is a 2x2 matrix...[[1,2],[2,3]]+[[1,2],[2,3]]=[[2,4],[4,6]]...right?
 
Sure. But in your problem, if I write it is as T(M)=A*M+M*A where A=[[1,2],[2,3]] you don't even know that A and M commute. So you can't write A*M+M*A as 2*A*M. And besides the matrix representation of T isn't 2x2. To do it directly a basis for 2x2 matrices is [[1,0],[0,0]], [[0,1],[0,0]], [[0,0],[1,0]] and [[0,0],[0,1]]. If you write M as a linear combination of those you can work out T as a 4x4 matrix. And then take the determinant of that 4x4. As I said, I'm not promising this is the easiest way. But it will work.
 
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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