1
$\begingroup$

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.

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

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.

$\endgroup$
  • $\begingroup$ The last part I would employ the hint that @rschwieb provided: a linear map is injective iff $\phi(x)=0 \implies x=0$ $\endgroup$ – Mathlete Mar 7 '13 at 21:08
  • $\begingroup$ That's very clear, thank you. $\endgroup$ – Mathlete Mar 7 '13 at 21:09
  • $\begingroup$ @Mathlete: You are welcome. And it sounds like you figured the last little bit with the hint given by rschwieb. $\endgroup$ – Thomas Mar 7 '13 at 21:11
  • $\begingroup$ Grr, but this is exactly what I said! :P No hard feelings though: congrats Thomas! $\endgroup$ – rschwieb Mar 7 '13 at 21:13
  • $\begingroup$ @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. $\endgroup$ – Thomas Mar 7 '13 at 22:22
2
$\begingroup$

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 $x=0$.)

$\endgroup$
  • $\begingroup$ Could you possibly go into a little more detail? $\endgroup$ – Mathlete Mar 7 '13 at 21:01
  • $\begingroup$ You are the horse at the water. I can do no more :/ $\endgroup$ – rschwieb Mar 7 '13 at 21:12
  • $\begingroup$ Then a thirsty horse I shall remain. $\endgroup$ – Mathlete Mar 7 '13 at 21:16
  • 1
    $\begingroup$ @Mathlete I don't know about that... you liked Thomas' answer, and it is just exactly what I wrote (with more symbols)! $\endgroup$ – rschwieb Mar 7 '13 at 22:04
1
$\begingroup$

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 $$c_1a_{1}+...+c_na_{n}=0$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.