Let X and Y be the vector spaces over R with dim(X) = n. Also, suppose there's a 1-1, onto lin. transformation, T, that maps X to Y. Show that dim(Y) = n.
Consider this approach (I did not do it but I am sure it works).
Let be a vector space over a field of dimension . Therefore, there exists a basis,
for .
Now, show that,
where is the bijective linear transformation between these vector spaces, that this set will form a basis for over .
This is mine 35th Post!!!
That they are linear independent is a result of:
suppose otherwise the there exist constants not all zeros such that:
but this would imply that:
which is a contradiction (as the 's are a basis and so linearly independent).
It is also obviouse that span , which is sufficient to show the result.
RonL
What CaptainBlank said is that when you have a linear transformation you can,
And,
.
So what he really did is combined them into a single expresssion,
(Actually I used the property of onto, here).
Now, because the map is one-to-one and a group homomorphism between these two vector spaces the kernel of the map (the nullspace) is trivial (zero vector).
Thus,
The fact that this forms a basis is true because the map is onto.