Say I have the following series of vectors $\left\{a,b,c,d,e\right\}$ with the following relations between them: $$a+b+c+d+e=0 \,\,\,\,\,;\,\,\,\,\, a \propto b\,\,\,\,\,;\,\,\,\,\,c=e\,\,$$ Then there exists three constraints among the five vectors so at most only two can be linearly independent. Now suppose we add another vector $k$ to the set, independent from the others. So now the dimensionality of the space spanned is $3$.
A possible spanning set is $\left\{k,a,e\right\}$ amongst many others but not e.g $\left\{k,a,b\right\}$.
Now construct the following four vectors $k+d, k, k+b+d+e$ and $k+a+b+d+e = k + d+ e + \rho b$, where $\rho$ is related to the proportionality constant between $a$ and $b$.
I find that these four vectors are linearly independent through solving the equation $\alpha k + \beta (k+d) + \gamma (k+b+d+e) + \delta (k+d+e+\rho b) = 0$ and finding $\alpha=\beta=\gamma=\delta=0$.
But why is this the case? Shouldn't they be linearly dependent because of the fact that $\text{dim}(\text{span} \left\{k,p_j\right\}) = 3$ which means I can create at most three linearly independent vectors and here I've used four?
Here $p_j$ is just any two appropriately chosen vectors out of the set $\left\{a,b,c,d,e\right\}$
Here is the working for solving of the constants: Vector by vector I get the following equations $$\alpha + \beta +\gamma +\delta=0$$$$\beta + \gamma + \delta = 0$$$$\gamma+ \rho \delta = 0$$$$\gamma + \delta = 0.$$ The first and the second imply $\alpha=0$ while the fourth with the second imply $\beta=0$. Then the third subtracted from the fourth imply $\delta(\rho-1)=0$ and here I took $\delta=0$ because $\rho=1$ would imply $a=0$. (as $a+b=\rho b$ and so $a = (\rho - 1)b$).
Note from the first condition that
$$ 0 = a+b+c+d+e = (1 + \rho) b+d+ 2 e $$ , using the other constraints. So you cannot have $b,d,e$ as independent vectors in the equation you set up for your four vectors:
$$ \alpha k + \beta (k+d) + \gamma (k+b+d+e) + \delta (k+d+e+\rho b) = 0 $$
Instead, replacing for example $e$ from the first condition, you have
$$ \alpha k + \beta (k+d) + \gamma (k+b+d-\frac12 ((1 + \rho) b+d)) + \delta (k+d+\rho b - \frac12 ((1 + \rho) b+d)) = 0 $$
So you have four variables $\alpha, \beta, \gamma, \delta$ but only three independent vectors $k,b,d$. So clearly $\alpha=\beta=\gamma=\delta=0$ is NOT the only solution. Therefore the four vectors are linearly dependent.