are you trying to show S is a subspace? Let's get that squared away first.
Let V be some vector space and S be a subset.
a. V=T(3,3) and S is the set of invertible linear transformations.
I know the V represents all linear transformations from R^3 -> R^3, and all linear transformations in S must be invertible, thus they must all be bijective. But I'm unsure how to go about this?
I know the three conditions that establish a subspace - S contains the 0 vector, S is closed under addition, and S is closed under scalar multiplication. Just not sure how to proceed.
I was unsure of what the 0 vector would be in T(2,2). Originally I thought it would be T(0), but I guess since every linear transformation can be expressed as T(x) = Ax, then the zero 3 by 3 matrix represents the zero matrix for T(2, 2). I assume this is true then for any vector space
T(m, n), where the zero vector is the n by m matrix.
But if we just say vector space T(m, n) contains all linear transformations of the form T:R^m -> R^n (regardless of whether they are invertible or not) then dim(T(m,n)) =mn, right? And since dim(R^n*m) = nm, can't one infer the dimensions are the same?
Note, when I write R^n*m I mean the vector space that contains all n by m matrices.
No, that's not true. In the vector space, T(n, n), of linear transformations (NOT necessarily invertible) from to , given a basis for , we can have take as basis the collection of transformations defined by if but for . That has dimension , not n. More generally, T(n, m), the set of linear transformations (NOT necessarily invertible) from to , given basis for and basis [itex]\{v_j\}[/itex] for , has basis the collection of transformations deined b with if and so has dimension nm.