Determining linear independence in $\mathbb{R}^3$ Let $\{\xi_k\}_{k=1}^4$ be a set of vectors in $\mathbb{R}^3$. If $\{\xi_1, \xi_2\}$ and $\{\xi_2, \xi_3, \xi_4\}$ are independent sets, and $\xi_1$ belongs to the span of $\{\xi_2, \xi_4\}$. Show that $\{\xi_k\}_{k=1}^3$ is linearly independent. 
Clearly, $\xi_1 = a_1\xi_2 + a_2\xi_4$ where $a_1\neq0$ due to the fact that $\xi_1$ and $\xi_2$ are linearly independent. Hence, $\xi_2 = a_1^{-1}\xi_1-\frac{a_2}{a_1}\xi_4$... How to proceed with the formalization? Because, it is quite clear that $\xi_1$ is linearly independent of $\xi_3$ and we know that $\{\xi_1, \xi_2\}$ is independent as well, so it follows that $\{\xi_k\}_{k=1}^3$ is linearly independent set. Is it rigorous enough?
Thank you! 
 A: Suppose that 
$$
\lambda_1\xi_1+\lambda_2\xi_2+\lambda_3\xi_3=0\tag{1}
$$
Writing $\xi_1=\alpha_2\xi_2+\alpha_4\xi_4$ then gives
$$
\lambda_1(\alpha_2\xi_2+\alpha_4\xi_4)+\lambda_2\xi_2+\lambda_3\xi_3=0
$$
Now, rearrange this equation to obtain
$$
(\lambda_1\alpha_2+\lambda_2)\xi_2+\lambda_3\xi_3+\lambda_1\alpha_4\xi_4=0\tag{2}
$$
Equation (2) and the linear independence of $\{\xi_2,\xi_3,\xi_4\}$ imply that $\lambda_3=0$. Equation (1) then implies that
$$
\lambda_1\xi_1+\lambda_2\xi_2=0\tag{3}
$$
Equation (3) and the linear independence of $\{\xi_1,\xi_2\}$ imply that $\lambda_1=\lambda_2=0$.
Hence $\{\xi_1,\xi_2,\xi_3\}$ is linearly independent.
Note that this argument avoids "dividing" by unknown scalars!
A: 
If $\{v_1,\dots,v_k\}$ is linearly independent in a vector space $V$ and $v\in V$, then $\{v_1,\dots,v_k,v\}$ is linearly independent if and only if $v$ doesn't belong to the span of $\{v_1,\dots,v_k\}$.

One direction is clear: if $v$ belongs to the span of $\{v_1,\dots,v_k\}$, the set $\{v_1,\dots,v_k,v\}$ is linearly dependent. Suppose $v$ doesn't belong to the span of $\{v_1,\dots,v_k\}$ and that
$$
\alpha_1v_1+\dots+\alpha_kv_k+\beta v=0
$$
Then $\beta=0$, otherwise $v$ would be a linear combination of $v_1,\dots,v_k$, against the hypothesis. Therefore $\beta=0$ and, by the linear independence of $\{v_1,\dots,v_k\}$, also $\alpha_1=\dots=\alpha_k=0$.

Thus you need to prove $\xi_3$ doesn't belong to the span of $\{\xi_1,\xi_2\}$. Suppose
$$
\xi_3=a\xi_1+b\xi_2
$$
Since you can write $\xi_1=c\xi_2+d\xi_4$, you get
$$
\xi_3=(ac+b)\xi_2+ad\xi_4
$$
Well, this is not possible, why?
A: hint
Start with a linear combination of vectors $1,2,3$ and set it equal to $0$ (vector). Now write vector $1$ as a linear combination of vectors $2$ and $4$. This gives you a linear combination of vectors $2,3$ and $4$ equal to the zero vector. Now use the independence of $2,3,4$ to find the coefficients.
