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After realizing my proof was incorrect, I tried it again.

Question:

If $L : V \rightarrow W$ is a linear mapping and ${L(v_1), . . . , L(v_n)}$ is linearly independent, then {v1, . . . , vn} is linearly inddependent.

I tried using a contradiction, but I feel like I made a mistake again (in a similar way).

1. Assume ${v_1,…,v_n}$ is linearly dependent. Then for the equation

$c_1v_1 + ... + c_n v_n = 0 $(1)

there exists a non-trivial solution.

2. Since zero vector maps to zero vectors in linear mappings, we have

$L(c_1 v_1 + .... + c_n v_n) = 0$ (2)

$c_1L(v_1) + ... + c_nL(v_n) = 0$ (3)

where at least one of $c_1,c_2 ... c_n$ is non zero.

This is a contradiction ...

But I feel like I made a mistake when going from equation 1 to equation 2 because since we put $"c_1v_1 + ... + c_n v_n$ into a different context (inside a linear mapping), the $c_1, c_2, ..., c_n$ can now hold different values and we are not guaranteed to have at least one c != 0.

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  • $\begingroup$ But $L$ is linear, so you can do that. $\endgroup$ – Randall Sep 20 '18 at 16:45
  • $\begingroup$ I'm talking about going from 1 to 2, not 2 to 3. $\endgroup$ – Tim Weah Sep 20 '18 at 16:54
  • $\begingroup$ Why can't you plug things in to functions? That's perfectly normal/acceptable/the point. $\endgroup$ – Randall Sep 20 '18 at 16:55
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    $\begingroup$ When you have $x=3$ and plug it into $f(x)=x^2$, does the $3$ change from the time you set it as $x$ to when you decided to plug it in? $f(3)=3^2=9$. $\endgroup$ – Randall Sep 20 '18 at 16:59
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    $\begingroup$ Tip: if something is easy, let it be easy! $\endgroup$ – Randall Sep 20 '18 at 17:08

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