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 $a = (\rho - 1)b$).

  • $\begingroup$ What is $p_j$? How do you get the dimension to be three? $\endgroup$
    – Dirk
    Jun 22 '17 at 10:57
  • $\begingroup$ @Bemte Ah sorry I made an edit in my post. dim =3 because of the three constraints among a,b,c,d and e makes at most 2 linearly independent and then I added k so it's three. (I think!) $\endgroup$
    – CAF
    Jun 22 '17 at 11:00
  • $\begingroup$ what does $a \propto b$ mean ? $\endgroup$ Jun 22 '17 at 11:01
  • $\begingroup$ @rapidracim, it probably means $a$ is proportional to $b$. $\endgroup$ Jun 22 '17 at 11:01
  • $\begingroup$ Yes, of course, by choosing only two vectors, you will get a dimenison of at most three... But you are given four vectors, one of them is $k$, not three, so I don't understand why you are using a total of three vectors now? $\endgroup$
    – Dirk
    Jun 22 '17 at 11:01

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.

  • $\begingroup$ Of course! Sorry, feel like I wasted everyone's time - thanks. $\endgroup$
    – CAF
    Jun 22 '17 at 11:19
  • $\begingroup$ So just to check my understanding, basically it means only three vectors at most constructed out of $\left\{k,a,b,c,d,e\right\}$ can be linearly independent? Four or more are always linearly dependent and three or less are always linearly independent? $\endgroup$
    – CAF
    Jun 22 '17 at 11:23
  • $\begingroup$ true, other than the last statement: "three or less are always linearly independent". Of course $a$ and $b$ are not linearly independent, and $c$ and $e$ are not linearly independent. $\endgroup$
    – Andreas
    Jun 22 '17 at 11:26
  • $\begingroup$ ah ok, so e.g $k, k+b,k+\rho b$ will be linearly dependent. $\endgroup$
    – CAF
    Jun 22 '17 at 11:33
  • $\begingroup$ yes......................... $\endgroup$
    – Andreas
    Jun 22 '17 at 11:38

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