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Thread: Lin Algebra (Tramsformations/Dimension Spaces)

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    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.
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    Quote Originally Posted by Bloden View Post
    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 35th Post!!!
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    Quote Originally Posted by ThePerfectHacker View Post
    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 35th 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
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    Quote Originally Posted by CaptainBlank View 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).
    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.
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    Quote Originally Posted by ThePerfectHacker View Post
    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

    RonL
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    Silly question, but what is the o's with a line through them? Obviously not the empty set.
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    Quote Originally Posted by Bloden View Post
    Silly question, but what is the o's with a line through them? Obviously not the empty set.
    It is the empty set.
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    Quote Originally Posted by Bloden View Post
    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
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    Quote Originally Posted by ThePerfectHacker View Post
    It is the empty set.
    ErrrHem ... cough cough ..

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

    RonL
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