Question about notation in Linear Algebra

BiGyElLoWhAt
Gold Member
Messages
1,637
Reaction score
138
Given vector spaces V, W, and a function T:V→W , state the two equations that the function T must satisfy to be a linear function.

Does T:V→W mean a function that maps vectors in V into W? Or what does this actually mean?
 
Physics news on Phys.org
pretty much, yeah. The domain of the function is V and the codomain is W.
 
  • Like
Likes 1 person
V or W could be the real numbers of course, or vector spaces of different dimensions.
 
T is a function that takes a vector ##v## in ##V## to a vector ##w## in ##W##. We are very used to the idea of function that take numbers as inputs. For instance, if ##f(x) = x^2 + 1##, then ##f## takes ##1## to ##2##. We denote this by ##f(1) = 2##.

So ##T: V \rightarrow W## means a function that takes v's to w's. I.e. ##T(v) = w##.

Now in vector spaces, any old function isn't that useful. We are specifically interested in linear transformations.

##T## is a linear transformation if the following two properties hold:

##1) \ T(v + v') = T(v) + T(v')## for all ##v, v' \in V##.
##2) \ T(cv) = cT(v)## for all ##v \in V## and scalars ##c##.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top