[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.