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True or false: If vectors $2u$, $3v$, and $4w$ are linearly independent, then $u$, $v$, and $w$, are also linearly independent. Explain.

This was a question on my test a second ago and I said false, because if $u$, $v$, or $w$ are the $0$ vector, then they are all linearly dependent by theorem $8$ in my textbook.

Because for constants $c_1, \dots, c_n$ if not all $c_i$'s are $0$, then

If $u$ is the zero vector then $c = 1$, then all other $c$ are $0$, and then the sum of vectors is $0$ --> linear dependent

did i do this right?

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    $\begingroup$ I think you made a mistake typing out the question statement. Is it supposed to say "linearly independent" twice or "linearly dependent" twice? Right now you have one of each, which doesn't make sense with the word "also". $\endgroup$ – user137731 Oct 16 '15 at 17:33
  • $\begingroup$ @Bye_World : Yes, that was annoying enough to make me fix it. :) $\endgroup$ – Disintegrating By Parts Oct 16 '15 at 19:07
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Suppose $\lambda_1 u + \lambda_2 v + \lambda_3 v = 0$. Then $$ \frac{\lambda_1}{2} (2u) + \frac{\lambda_2}{3}(3v) + \frac{\lambda_3}{4}(4w) = 0. $$ But that implies $\lambda_1/2 = \lambda_2/3 = \lambda_3/4 = 0$, since $2u,3v,4w$ are linearly independent. So each $\lambda_i$ is also 0. Therefore if $2u, 3v,4w$ are linearly independent, then so are $u,v,$ and $w$.

Also, if $2u,3v,$ and $4w$ are linearly dependent, then we can use the same trick as the above to show that $u,v,$ and $w$ are linearly dependent. Just take a relation $\lambda_1(2u) + \lambda_2(3v) + \lambda_3(4w) = 0$. Then $(2\lambda_1)u + (3 \lambda_2)v + (4\lambda_3) w = 0$. At least one of the $\lambda_i$ must be non-zero, so one of $2\lambda_1,...,4\lambda_3$ is non-zero.

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If the vectors $2u$, $3v$, and $4w$ are linearly independent, then none of $u$, $v$, or $w$ is the zero vector. Any set of vectors containing the zero vector is linearly dependent for the reason you described, but given the linear independence of $\{2u, 3v, 4w\}$, that case is ruled out.

In general, multiplying each of the vectors in a set by non-zero scalars does nothing to change their linear independence (or lack thereof). If there is a non-trivial linear combination of the original vectors that adds to zero, then there is a non-trivial linear combination of the new vectors that adds to zero (simply divide each coefficient by the appropriate scalar).

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