Prove: the set of 3x3 symmetric matrices is a vector space and find its dimension. Well in class my prof has done this question, but I still don't quite get it.. Ok, first off, I need to prove that it's a vector space. The easy way is probably to prove that it contains the zero space and is closed under the two operations. Certainly it contains the zero vector, which I understand; but when my prof went for the second part, he wrote something in this fashion: [a d e]___[1 0 0]____[0 0 0] [d b f] = a [0 0 0] + b [0 1 0] + c [...] + d [...] + e [...] + f [...] => dimension of 6. [e f c]___[0 0 0]____[0 0 0] (sorry I have to use the underscore to line them up properly..) How is that closed under the two operations? I am not getting it completely.. Also, how they add up to the original matrix certainly makes sense to me, but how he devided them up in this way sort of puzzles me...Since every matrix is reducible, so this one should be as well. Since the dimension of the number of elements in a basis, then shouldn't this have a basis of max. 3? I'm pretty sure I am not thinking correctly, but someone please straighten this stuff out for me please!