Are $\vec{v}$ and $\vec{w}$ linearly independent? Am I correct in saying that $\vec{v}=\begin{pmatrix}1\\2\\0\end{pmatrix}$ and $\vec{w}=\begin{pmatrix}2\\4a\\a-1\end{pmatrix}$ are linearly independent $\forall a\in\mathbb R$ /{1}? It seems to me that the only way the second vector can be a multiple of the first is if $a=1$. How could I prove this? I tried something but I don't know if that is correct:
1) $\vec{v} ,\vec{w}$ are linearly independent if and only if the only solution to $\lambda\vec{v}+\mu\vec{w}=\vec{0}$ is $\lambda=\mu=0$
2) $\lambda\begin{pmatrix}1\\2\\0\end{pmatrix}+\mu\begin{pmatrix}2\\4a\\a-1\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$
$\Rightarrow \lambda+2\mu=0$
$2\lambda+4a\mu=0$
$\mu a-\mu=0$
$\Rightarrow a=1$
I am not sure why I am getting $a=1$ because isn't that the solution where they are linearly dependent?
 A: *

*If $a=1$ then $w=2v$ so they are linearly dependent.

*If $a\ne1$ then $\mu a-\mu=0\iff \mu =0$ and then $\lambda+2\mu=0\implies \lambda=0$ so $\mu=\lambda=0$ hence the two vectors $v$ and $w$ are linearly independent.
A: To summarize what you've said: $\vec{v} ,\vec{w}$ are linearly independent if and only if the only solution to $\lambda\vec{v}+\mu\vec{w}=\vec{0}$ is $\lambda=\mu=0$.
For this particular choice of $\vec v,\vec w$, we have $\lambda,\mu$ are a solution to $\lambda\vec{v}+\mu\vec{w}=\vec{0}$ if and only if $\lambda + 2 \mu = 0$ and
$$
\mu a - \mu = 0 \iff\\
\mu(a - 1) = 0
$$ 
That is, if $a \neq 0$, then the only solution is $\mu = \lambda = 0$.  However, if $a = 1$, then it suffices to have $\lambda + 2 \mu = 0$, giving us infinitely many solutions.  So, when $a = 1$, the two vectors are linearly dependent.
A: $$\begin{pmatrix}2\\4a\\a-1\end{pmatrix}=k\begin{pmatrix}1\\2\\0\end{pmatrix}\iff \begin{cases}k=1&,\;\;\text{line}\;1\\{}\\k=2a&,\;\;\text{line}\;2\\{}\\a=-1&,\;\;\text{line}\;3\end{cases}$$
But then
$$(1)+(2)\;\;\implies\;\;1=k=2a\implies a=\frac12$$
which is inconsistent with third line. Thus the vectors are linearly independent.
