Thread: Lin 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.

Originally Posted by Bloden

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!!!

Originally Posted by ThePerfectHacker

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

Originally Posted by CaptainBlank

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).

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.

Originally Posted by ThePerfectHacker

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.

And I thought I aaid that

RonL

Silly question, but what is the o's with a line through them? Obviously not the empty set.

Originally Posted by Bloden

Silly question, but what is the o's with a line through them? Obviously not the empty set.

It is the empty set.

Originally Posted by Bloden

Silly question, but what is the o's with a line through them? Obviously not the empty set.

Pay no attention, they are the Greek letter phi, it denote the one-one and
onto linear transformation which I see the question uses T for.

RonL

Originally Posted by ThePerfectHacker

It is the empty set.

ErrrHem ... cough cough ..

(He is asking about the 's)

RonL