0

I came across this demonstration Proving a linear transformation is unique. I get how "peek-a-boo" proved this theorem but I don't understand when the uniqueness of the scalars comes handy. Wouldn't a generating set be enough to make the same proof?

J. W. Tanner
  • 63,683
  • 4
  • 43
  • 88
Piedino
  • 11
  • 1
    Show us what you mean by this question. – Ted Shifrin Feb 04 '24 at 23:30
  • A bit of a translation for those of us who use alternate terminology: Piedino is wondering why it's been proven that the linear transformations are uniquely defined on bases, rather than just on spanning sets, and doesn't understand where linear independence is being used. – Theo Bendit Feb 05 '24 at 00:21

1 Answers1

5

It is in fact correct that if two linear maps agree on a generating set then they agree on the whole of the space. The fact the scalar coefficients exist here is sufficient, uniqueness is not used or necessary.

However given a specified action on a generating set, there may not necessarily exist a linear map extending this to the whole space (the outputs of the generating set must satisfy all of the same relations as the generating set did, otherwise the information is inconsistent - note bases satisfy no non-trivial linear relations so this is not an issue here). If such a map does happen to exist, it is still certainly unique by the previous discussion however.

In light of this, we often restrict to specifying actions on bases rather than general generating sets to yield a (well-defined) linear map.

slowlight
  • 375