# Lin Algebra (Tramsformations/Dimension Spaces)

• Nov 29th 2006, 12:01 PM
Bloden
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.
• Nov 29th 2006, 12:29 PM
ThePerfectHacker
Quote:

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 $\displaystyle X$ 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 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
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!!!

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 PM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlank
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).

What CaptainBlank 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 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
What CaptainBlank 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.

And I thought I aaid that:rolleyes:

RonL
• Nov 30th 2006, 01:36 AM
Bloden
Silly question, but what is the o's with a line through them? Obviously not the empty set.
• Nov 30th 2006, 07:39 AM
ThePerfectHacker
Quote:

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.
• Nov 30th 2006, 11:56 AM
CaptainBlack
Quote:

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
• Nov 30th 2006, 11:58 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
It is the empty set.

ErrrHem ... cough cough ..

(He is asking about the $\displaystyle \phi$'s)

RonL