Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I found the following theorem in "Friedberg-Lienar algebra 4ed".

" Let $~~V,~W~$ be vector space over field F.

Let $~ \varphi ~: V \to W ~~$ be $~~$ isomorphism .

Then, For any $~Q \subset V$ , $~~$Span$( \varphi(Q) ) $= $~ \varphi ($Span(Q)$)$ . "

That can be generalized where $~ \varphi ~$ is any linear map?

My proof is as follows.

Pf) Take any $~Q \subset V$. ( I use the Einstein convention.)

$( \Rightarrow )$ :$~~~$trivial.

$~Q \subset Span(Q)$ $~ \Rightarrow $ $~\varphi(Q)$ $\subset \varphi ($Span(Q)$)$ $~~ \Rightarrow $ $~Span(\varphi(Q))$ $< \varphi ($Span(Q)$)$.

$( \Leftarrow )$ :

$\Bbb D $ $\in $ $\varphi ($Span(Q)$)$ $~ \Rightarrow $ $\exists x_1~,...,x_k \in F$ and $~\Bbb G_1~,...,\Bbb G_k \in Q$ such that $\Bbb D= x_j \varphi(\Bbb G_j) $

$ \Rightarrow $ $\Bbb D= x_j \varphi(\Bbb G_j) \in $ $~Span(\varphi(Q))~~$ (Because $ \varphi(\Bbb G_1),~...~, \varphi(\Bbb G_k) \in Span$$(\varphi(Q))$.

Is that right?

share|improve this question
    
Friedberg is written in broken English? What's the Einstein convention? –  Gerry Myerson Aug 6 '13 at 9:28
    
@GerryMyerson/ en.wikipedia.org/wiki/Einstein_notation –  Chris kim Aug 7 '13 at 7:25
add comment

1 Answer 1

If $v=\sum_i c_iq_i$ (with $v\in V, c_i\in F, q_i\in Q$) is any (by definition finite) linear combination of vectors in$~Q$, then by linearity of $\varphi$ one has $$ \varphi(v) = \varphi\left(\sum_i c_iq_i\right) = \sum_i c_i\varphi(q_i). $$ Therefore a vector (in $W$) that can be written in the form of the first expression if and only if it can be written in the form of the final expression; in other words $\def\span{\operatorname{span}}\varphi(\span(Q))=\span(\varphi(Q))$.

Clearly only linearity of $\varphi$ is needed, nothing about being injective or surjective (and the spaces do not have to be finite dimensional). I think the proof you gave will boil down to the above equality if you consider it closely, but the formulation you gave makes the elementary essence of the argument less transparent.

share|improve this answer
    
/ I didn't consider the "injective" and "surjective". :) So,... my proof is correct? –  Chris kim Aug 7 '13 at 1:40
    
@Chriskim Your proof has no real errors (but the Einstein convention requires to write one upper and one lower index if you want to implicitly sum over them). However I think there is a lot to be improved, as it makes a lot of "noise" about stuff having little to do with the actual proof (not to mention the gratuitous use of "reserved" symbols $\Bbb D,\Bbb G$) while the true action is passed over in silence: the replacement of $\varphi(Q)$ by $Span(\varphi(Q))$ in the "trivial" part (needs justification), and writing $x_j\varphi(G_j)$ instead of $\varphi(\sum_j x_jG_j)$ in the other part. –  Marc van Leeuwen Aug 7 '13 at 6:44
    
@Mark van Leeuwen/ Ah..! Ok! I understand! Thank you so much! :) –  Chris kim Aug 7 '13 at 7:22
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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