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?
Best Answer
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.