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This is an extension of this question. To rehash the main ideas, we have vectors $(0,0,0,0,0,0)$, $(a,a,a,0,0,0)$, $(b,b,b,b,b,b)$, $(c^0,c^1,c^2,c^0,c^1,c^2)$ and $(0,d,2d,0,0,0)$ in a linear algebra system.

I'm hoping that we can create a (6-dimensional) basis for the system using an additional vector: $(0,-e,0,0,0,2e)$. I obtain this vector using the other vectors, knowing there are limits to the size of each value in the vectors. I get $(c^0,c^1-d,c^2-2d,c^0,c^1,c^2)$ by adding two of the original vectors. Then I make $c$ very large. The idea is that I then set $2d=c^2$, and that cancels out one of the values in the new vector, so that I'm left with $(c^0,c^1-d,0,c^0,c^1,c^2)$. Since $c$ and $d$ are very large, it follows from a little bit of thought that this new vector is approximately $(0,0-d,0,0,0,c^2)$. Again, from $2d=c^2$, we have this roughly equal to $(0,-d,0,0,0,2d)$. Does this help to form a 6-dimensional system? In other words, can I determine all 6 indices from these vectors?

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The null vector isn't really a part of your system, and you're not really talking about a 'basis' because one of your vectors - specifically, {1, c, c^2, 1, c, c^2} - isn't linear, so you're not taking linear combinations. –  Steven Stadnicki Nov 16 '11 at 0:14
    
@Steven Stadnicki: I was afraid of that. I can also get $\{0,c^0-2d-c^1,0,-2c^0-2d,-c^0-2d-c^1,0\}$ by similar techniques. I get this by subtracting from a vector of all "$c^2$". –  Matt Groff Nov 16 '11 at 0:14
    
A quick code gave the result as $\left(\begin{array}{cccccc} -\frac{1}{c - 1} & \frac{2}{c - 1} & -\frac{1}{c - 1} & -1 & 1 & 0\\ -\frac{c + 1}{c - 1} & \frac{2\, \left(c + 1\right)}{c - 1} & -\frac{c + 1}{c - 1} & -1 & 0 & 1 \end{array}\right)^T $ as a possible answer. –  user13838 Nov 16 '11 at 0:24
    
@Steven Stadnicki: I should add that I'm getting numbers for all of these values. They all represent values that sum up to one sum. So when I take linear combinations, I'm really taking them from a single-valued system. I guess that I may still be in trouble trying to solve the system, though. That's whay I'd really like to know! –  Matt Groff Nov 16 '11 at 0:27
    
Please note that curly brackets, $\{$ and $\}$, are used to denote sets, not vectors. In a set, order does not matter, so to most mathematicians, $\{1,2,3\}$ will be the same as $\{3,2,1\}$. If you want to avoid confusion, you should use some other delimiter, preferably $($ and $)$. –  Arturo Magidin Nov 16 '11 at 18:28
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1 Answer

up vote 3 down vote accepted

As before, you have the vector $(0,1,2,0,0,0)$, the vector $(1,1,1,0,0,0)$, and vectors of the form $(1,c,c^2,1,c,c^2)$, which give you three dimensions by taking $c=1$, $c=2$, and $c=3$, say (you cannot get more than three linearly independent such vectors, and three are linearly independent if and only if the three values of $c$ are pairwise distinct. So the linear span of all these vectors you already had has a basis given by $$(0,1,2,0,0,0),\quad (1,1,1,0,0,0),\quad (1,1,1,1,1,1),\quad (1,2,4,1,2,4),\quad (1,3,9, 1,3,9).$$

You want to know if you can find a value of $e$ for which $(0,-e,0,0,0,2e)$ is linearly independent of these vectors. That's the same as asking if $(0,1,0,0,0,-2)$ is linearly independent from the other five.

The answer is that, no, they are not linearly independent: the vectors $(1,1,1,1,1,1)$, $(1,2,4,1,2,4)$, and $(1,3,9,1,3,9)$ are linearly independent, and they are the only ones that contribute to the last three components. So you can find a linear combination of them (in fact, a unique linear combination) such that $$(0,0,-2,0,0,-2) = \alpha(1,1,1,1,1,1) + \beta(1,2,4,1,2,4) + \gamma(1,3,9,1,3,9).$$ A bit of work shows that $\alpha=-1$, $\beta=2$, and $\gamma=-1$ will do.

But now add the vector $(0,1,2,0,0,0)$ to this linear combination to obtain $(0,1,0,0,0,-2)$. That is, every vector of the form $(0,=-e,0,0,0,2e)$ is already a linear combination of the vectors you have, so you are still "locked in" in the same 5-dimensional subspace of $\mathbb{R}^6$.

So, no, even with these additional vectors you cannot obtain all vectors in $\mathbb{R}^6$ via linear combinations of the vectors you "have".

(Note that if you "create" a vector using other vectors you already "have", and all the operations are linear, then the new vectors cannot enlarge the linear span of your set.)

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