Thread: Domain and Range of a Linear Transformation

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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.

yes, but can you justify your answer?

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:R²-->R², the domain is always going to be R². the range "might" be R², or it might be a subspace of R².

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:Rⁿ-->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) = Rⁿ.

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 Rⁿ onto Rⁿ...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 Rⁿ (this has n elements, one for each coordinate position, so dim(Rⁿ) = 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.