0
$\begingroup$

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

$\endgroup$
11
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.