yes, but can you justify your answer?
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In problems 3c and 3d would it be correct of me to say that the domain and range of T are R^2? I don't understand how to answer this sort of question since there are no examples in my book.
Well, the domain of the linear transformation T(x) = Ax = b would be the span of x right? The range of T would be the span of b.
I can't really justify it beyond looking at x and b and seeing that they have two rows so thus they are in R^2. I'm not really confident in this answer because x or b could potentially not span all of R^2 and R^2 would not be a correct answer (I think). Would saying that the domain and range are span(x) be a better answer?
the domain of a linear transformation is the space on which it is defined. linear transformations aren't like real-valued functions (in general) there typically aren't "points where they're undefined".
the range, on the other hand, is an entirely different matter. it's typically a LOT smaller than the co-domain (the target space) unless: the linear transformation is ONTO.
for a linear transformation T:R^{2}-->R^{2}, the domain is always going to be R^{2}. the range "might" be R^{2}, or it might be a subspace of R^{2}.
if the transformation T is (or is represented by) a matrix A, then we can say certain things about T based just on the matrix A. we can do the general case:
T:R^{n}-->R^{m} without much extra trouble. in this case, we get an mxn matrix for A.
range(T) = column space of A. this has dimension rank(A), which is going to be less than or equal to m. if rank(A) = m, we say T (or A) is onto. this space is spanned by the columns of A, but we might not need "all" of them (rank(A) could be < m).
domain(T) = R^{n}.
of particular interest is the case where m = n. in this case, when rank(A) = n, we say T (or A) is invertible. rank and range are closely related: the rank is the DIMENSION of the range.
there is a "push-me pull-you" relationship between the rank of a matrix and the dimension of its null space, the bigger the rank, the smaller the null space. at the extremes you have the 0-matrix (which has 0 rank, and n-dimensional null space), and the invertible matrices (which map R^{n} onto R^{n}...the simplest example is the identity transformation (with the matrix I): it has a null-space of 0 dimension (just {0}, which has no basis) and is of rank n).
it is bad thinking to think of the domain as span(x) and the domain as span(b) for the equation Ax = b. the span of a single vector is just one-dimensional (or 0-dimensional if that vector is the 0-vector).
how do we find out what the dimension of span(S) is, for a set S of vectors? we produce a basis. when in doubt, you can often use this basis:
{(1,0,...,0), (0,1,0,...,0),...,(0,...,0,1)} for R^{n} (this has n elements, one for each coordinate position, so dim(R^{n}) = n).
when you're given a finite set X of k vectors, you know straight off the bat that dim(span(X)) ≤ k, but it may not actually BE k. some of the vectors might be linearly dependent, and therefore redundant. for example:
span({(1,1),(3,3)} = span({(1,1)}, since (3,3) is linearly dependent on (1,1): 3(1,1) + (-1)(3,3) = (0,0), but this is not a 0-linear combination.