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Let $\phi : A \rightarrow B$ be a linear map and $a_{1},...,a_{n} \in A$

i) If $a_{1},...,a_{n}$ span $A$ and $\phi$ is surjective, show that $\phi(a_{1}),...\phi(a_{n})$ span $B$.
Is this also true when $\phi$ is not surjective?

ii) If $a_{1},...,a_{n}$ are linearly independent and $\phi$ is injective, show that $\phi(a_{1}),...\phi(a_{n})$ are linearly independent. Is this also true when $\phi$ is not injective?

I don't know where to begin. Any help would be appreciated.

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what are you trying ? –  Maisam Hedyelloo Mar 7 '13 at 20:50
    
I don't really know where to start. I'm trying to look over my definitions. –  Mathlete Mar 7 '13 at 20:52
1  
This follows directly from the definitions, without any trick. Start by writing down your assumptions in more details. –  1015 Mar 7 '13 at 20:52

3 Answers 3

up vote 1 down vote accepted

1) So you want to show that $\phi(a_i)$ span $B$. Let $b\in B$. Then there is a $$ a = c_1a_1 + \dots + c_na_n \in A $$ such that $\phi(a) = b$ (because $phi$ is surjective. But that means that $$ b = c_1\phi(a_1) + \dots + c_n\phi(b_n). $$

2) Say that $$ c_1\phi(a_1) + \dots+c_n\phi(a_n) = 0. $$ Then $$ \phi(c_1a_1 + \dots + c_na_n) = 0 $$

and you can probably finish it from here.

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The last part I would employ the hint that @rschwieb provided: a linear map is injective iff $\phi(x)=0 \implies x=0$ –  Mathlete Mar 7 '13 at 21:08
    
That's very clear, thank you. –  Mathlete Mar 7 '13 at 21:09
    
@Mathlete: You are welcome. And it sounds like you figured the last little bit with the hint given by rschwieb. –  Thomas Mar 7 '13 at 21:11
    
Grr, but this is exactly what I said! :P No hard feelings though: congrats Thomas! –  rschwieb Mar 7 '13 at 21:13
    
@rschwieb: I upvoted you answer :) I actually hadn't read it before I posted my answer, but yes you are saying the same as me. –  Thomas Mar 7 '13 at 22:22

Hint for 1) For a generic element $b\in B$, find an element of $A$ in terms of the generating set that maps to $b$. Then think about the linearity of $\phi$.

Hint for 2) Look at a linear combination of the $\phi(a_i)$ equal to zero and apply linearity of $\phi$. (It is especially helpful for this half to know that a linear map is injective iff $\phi(x)=0$ implies $x=0$.)

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Could you possibly go into a little more detail? –  Mathlete Mar 7 '13 at 21:01
    
You are the horse at the water. I can do no more :/ –  rschwieb Mar 7 '13 at 21:12
    
Then a thirsty horse I shall remain. –  Mathlete Mar 7 '13 at 21:16
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@Mathlete I don't know about that... you liked Thomas' answer, and it is just exactly what I wrote (with more symbols)! –  rschwieb Mar 7 '13 at 22:04

for a $<a_1,a_2,...,a_n>=A $ we want show that $$B=<\phi (a_{1}),...,\phi(a_{n})>$$ clearly $$<\phi (a_{1}),...,\phi(a_{n})>\subset B$$ now prove$$ B\subset <\phi (a_{1}),...,\phi(a_{n})> $$ let $y\in B$ $ \exists x\in A$ such that $\phi(x) =y$ where $$x\in<a_1,a_2,...,a_n> ,\exists c_i \ s.t \quad x=\sum_{i=o}^nc_ix_i$$ hence $$\phi(x) =\sum_{i=o}^nc_i\phi(x_i)$$
$$y\in<\phi (a_{1}),...,\phi(a_{n})>$$ for b $c_1\phi (a_{1})+...+c_n\phi(a_{n})$=0then $$\phi (c_1a_{1}+...+c_na_{n})=0$$ therefore $$c_1a_{1}+...+c_na_{n}=0$$

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