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

Fernando Revilla

Well-known member
MHB Math Helper
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.

Fernando Revilla

Well-known member
MHB Math Helper
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 Linear and Abstract Algebra section.