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$L : \mathbb{R}^n \longrightarrow \mathbb{R}^n$ is a linear mapping. Prove if $\textrm{span}\{v_1,\dots , v_n\}=\mathbb{R}^n$ and $\{L(v_1), \dots , L(v_n)\}$ is linearly independent set in $\mathbb{R}^n$, then $\ker(L) = \{0\}$.

What I have so far: If $\textrm{span}\{v_1,\dots , v_n\}=\mathbb{R}^n$, any vector $x \in \mathbb{R}^n$ can be written as $c_1v_1 + \dots + c_nv_n$ for some $c_1,\dots,c_n$ in the reals.

Subbing this vector $x$ into $L$ gives $L(x) = L(c_1v_1 +\dots+ c_nv_n) = c_1L(v_1) +\dots + c_nL(v_n)$. This is where I get confused. Doesn't $c_1L(v_1) +\dots + c_nL(v_n)$ being linearly independent imply any vector in $\mathbb{R}^n$ would result in the zero vector? Did I do something wrong?

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    $\begingroup$ $c_1L(v_1)+c_2L(v_2)+........c_nL(v_n) =0$ is said to be linearly independent $\iff$ $c_1=c_2=c_3=.......=c_n=0$ $\endgroup$ – user210387 Jun 16 '15 at 15:23
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Suppose $v\in\ker(L)$, that is, $L(v)=0$. Write $v=c_1v_1+\dots+c_nv_n$, which is possible because $\{v_1,\dots,v_n\}$ spans $\mathbb{R}^n$. Then $$ 0=L(v)=L(c_1v_1+\dots+c_nv_n)=c_1L(v_1)+\dots+c_nL(v_n). $$ Now the linear independence of $\{L(v_1),\dots,L(v_n)\}$ tells you that $$ c_1=0,\dots, c_n=0 $$ hence that $v=0$.

Recall: $\{w_1,\dots,w_n\}$ is linearly independent if and only if $c_1w_1+\dots+c_nw_n=0$ implies $c_1=0,\dots,c_n=0$. Just apply the definitions.

This is actually a special case of the rank-nullity theorem: the set $\{v_1,\dots,v_n\}$ is a basis of $\mathbb{R}^n$ (being a spanning set with $n$ elements) and so $\{L(v_1),\dots,L(v_n)\}$ is a spanning set for the range of $L$. Since this is by hypothesis linearly independent, the rank of $L$ is $n$ and the rank-nullity theorem gives $\dim\ker(T)=n-n=0$.

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  • $\begingroup$ Even I wrote a proof for it but I don't get how you imply that $c_1=0,\dots, c_n=0$ then $v=0$@egreg $\endgroup$ – user210387 Jun 16 '15 at 15:58
  • $\begingroup$ @Rememberme $v=c_1v_1+\dots+c_nv_n=0v_1+\dots+0v_n=0$. $\endgroup$ – egreg Jun 16 '15 at 16:00
  • $\begingroup$ Okay okay.. Ah I should have seen that $\endgroup$ – user210387 Jun 16 '15 at 16:01
  • $\begingroup$ Well I used rank nullity $\endgroup$ – user210387 Jun 16 '15 at 16:01
  • $\begingroup$ Right, thats where I also get lost. if c1=0,…,cn=0 for c1L(v1)+⋯+cnL(vn) = 0. can't any vector in R^2 satisfy this equation? $\endgroup$ – Filip Jun 16 '15 at 16:03

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