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.

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- Nov 29th 2006, 01:01 PMBlodenLin Algebra (Tramsformations/Dimension Spaces)
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.

- Nov 29th 2006, 01:29 PMThePerfectHacker
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 35:):)th Post!!! - Nov 29th 2006, 02:04 PMCaptainBlack
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 - Nov 29th 2006, 02:12 PMThePerfectHacker
What Captain

**Blank**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. - Nov 29th 2006, 02:15 PMCaptainBlack
- Nov 30th 2006, 02:36 AMBloden
Silly question, but what is the o's with a line through them? Obviously not the empty set.

- Nov 30th 2006, 08:39 AMThePerfectHacker
- Nov 30th 2006, 12:56 PMCaptainBlack
- Nov 30th 2006, 12:58 PMCaptainBlack