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, 12:01 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, 12:29 PMThePerfectHacker
Consider this approach (I did not do it but I am sure it works).

Let $\displaystyle X$ be a vector space over a field $\displaystyle R$ of dimension $\displaystyle n$. Therefore, there exists a basis,

$\displaystyle \{ x_1,x_2,...,x_n\}$ for $\displaystyle X$.

Now, show that,

$\displaystyle \{ \phi(x_1), \phi(x_2),...,\phi(x_n) \}$ where $\displaystyle \phi$ is the bijective linear transformation between these vector spaces, that this set will form a basis for $\displaystyle Y$ over $\displaystyle R$.

This is mine 35:):)th Post!!! - Nov 29th 2006, 01:04 PMCaptainBlack
That they are linear independent is a result of:

suppose otherwise the there exist constants $\displaystyle a_1,\ ..,\ a_n$ not all zeros such that:

$\displaystyle

a_1 \phi(x_1)+ ...+ a_n \phi(x_n)=\bold{0}

$

but this would imply that:

$\displaystyle

a_1 x_1+ ...+ a_n x_n=\bold{0}

$

which is a contradiction (as the $\displaystyle x$'s are a basis and so linearly independent).

It is also obviouse that $\displaystyle \{ \phi(x_1), \phi(x_2),...,\phi(x_n) \}$ span $\displaystyle Y$, which is sufficient to show the result.

RonL - Nov 29th 2006, 01:12 PMThePerfectHacker
What Captain

**Blank**said is that when you have a linear transformation you can,

$\displaystyle \alpha \phi (x)=\phi(\alpha x)$

And,

$\displaystyle \phi (x+y)=\phi(x)+\phi(y)$.

So what he really did is combined them into a single expresssion,

$\displaystyle \phi (a'_1 x_1+...+a'_n x_n)=0$

(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,

$\displaystyle a'_1 x_1+...+a'_n x_n=0$

The fact that this forms a basis is true because the map is onto. - Nov 29th 2006, 01:15 PMCaptainBlack
- Nov 30th 2006, 01:36 AMBloden
Silly question, but what is the o's with a line through them? Obviously not the empty set.

- Nov 30th 2006, 07:39 AMThePerfectHacker
- Nov 30th 2006, 11:56 AMCaptainBlack
- Nov 30th 2006, 11:58 AMCaptainBlack