[quote=crystal-maiden;285899]T: R^3 ----> W , where W is the vector space of all symmetric 2x2 matrices
T [ a b c ] =
[ a+b+c b-2c ]
[ b-2c a-c ]
I am new to abstract algebra, but I have a few thoughts/questions.
1) A basis for the vector space of symmetric matrices isBut I think I know it is not "onto" because T of R^3 cannot possibly cover all of the range of M22. A side-question on a little notation problem, is the space of 2x2 matrices M22 ?
Therefore it has dimension 3, so it is possible to create a 1-to-1 and onto map, T. For instance,
This T will hit every symmetric matrix (onto), and every element in maps to a unique symmetric matrix.
This shows it is possible for a T. But your question is asking about your T not any T.
2) I do not know how you concluded your values above.I'm not sure how to show it is 1 to 1. After being transformed, I see that a = c and b = 2c and 2a = -b.
3) Consider the symmetric matrix:
Your transformation is going to take into the symmetric matrix:
This gives us three equations, namely:
we can express this in terms of matrices
if you row-reduce this last matrix you will see that it has full rank.
First, this means that for any (any symmetric matrix) there is a unique solution for (an element in ). Thus, T is onto.
Second, the full rank shows that each in maps to a unique in the 2x2 symmetric matrix space. Thus, T is one-to-one.
Like I said, i am new to abstract algebra, I hope this at least gets you started.