MHB Cameron V 's question at Yahoo Answers (Equality of linear maps)

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The discussion revolves around proving that if two linear transformations L1 and L2 yield the same outputs for a basis of a vector space V, then L1 must equal L2. The proof begins by establishing that if L1 equals L2, then their outputs for each basis vector are trivially equal. Conversely, if L1 and L2 produce the same results for all basis vectors, a generic vector x in V can be expressed as a linear combination of these basis vectors. By applying the linear transformations to this combination, it is shown that L1(x) equals L2(x), leading to the conclusion that L1 equals L2. The thread encourages further questions to be posted in a dedicated math help forum.
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Here is the question:

Prove:
for each i = 1, 2, ... n
than L1(vi) = L2(vi)
only if L1 = L2

I can this of the concept in my head and I think I understand it but I am having trouble actually putting the proof on paper. Any help is appreciated.
Thanks

Here is a link to the question:

{v1, v2, ... , vn} is a basis for V. L1 and L2 are two linear transformations mapping V into a vectorspace W.? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hello Cameron V,

If $L_1=L_2$, trivially $L_1(v_i)=L_2(v_i)$ for all $i=1,\ldots,n$. On the other hand if $L_1(v_i)=L_2(v_i)$ for all $i=1,\ldots,n$, choose a generic $x\in V$. As $\{v_1,\ldots,v_n\}$ is a basis of $V$, $x=\alpha_1+\ldots +\alpha_n v_n$ for some scalars $\alpha_1,\ldots,\alpha_n$. Then, for all $x\in V$: $$\begin{aligned}L_1(x)&=L_1(\alpha_1v_1+\ldots +\alpha_n v_n)\\&=\alpha_1L_1(v_1)+\ldots +\alpha_n L_1(v_n)\\&=\alpha_1L_2(v_1)+\ldots +\alpha_n L_2(v_n)\\&=L_2(\alpha_1v_1+\ldots +\alpha_n v_n)\\&=L_2(x)\\&\Rightarrow L_1=L_2\end{aligned}$$ If you have further questions, you can post them in the http://www.mathhelpboards.com/f14/ section.
 
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