Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm researching a potential algorithm, and I'm hoping that someone can verify my calculations.

I have sets of vectors in $\mathbb{R}^6$ that I can use. They have a corresponding value associated with them, but this relation is not necessarily a simple one. I'd like to find the value associated with the vector $(0,0,1,0,0,0)$. So I'm wondering if I can perform linear algebra, using the vectors that I can create, to do so.

One vector I can create is $(a,a,a,0,0,0)$ for any real $a$. A second is $(b,b,b,b,b,b)$.

I can also create vectors of the form $(c^0,c^1,c^2,c^0,c^1,c^2)$ for some real number $c$, where $c^k$ is just simply $c$ taken to the $k$th power. A second form that I can create is $(0, d, 2d, 0, 0, 0)$ for real $d$.

I am wondering if these vectors form a complete basis for $\mathbb{R}^6$.

In case it helps, I can use as many vectors as I want, adding and/or subtracting them, as long as they are of the forms above.

My Question

What I really want to know is, can I find the corresponding value for $(0,0,1,0,0,0)$?

share|improve this question
    
What does $c^{k}$ mean? And can you expand more on your explanation. Thanks. –  Dan Donnelly Apr 2 '11 at 2:00
    
I've edited the question, so hopefully it makes more sense now. Please let me know if anything is still unclear! Thanks. –  Matt Groff Apr 2 '11 at 2:14
    
I'm a bit confused about your question are you asking whether or not you can form a basis for R^6 out of vectors of the form you described or whether or not {0,0,1,0,0,0} is in a span of vectors of the form you described or something else? –  WWright Apr 2 '11 at 2:16
1  
@Matt: curly brackets are used to denote sets, not vectors; in a set, order and repetitions are immaterial. So $\{a,a,a,0,0,0\} = \{a,0\} = \{0,a,a\} = \{0,0,0,a,a,a\}$. If you mean vectors (in the usual "ordered tuple" notion), then you should use parentheses, not curly brackets. –  Arturo Magidin Apr 2 '11 at 2:19
    
@WWright: I really only want to know if $(0,0,1,0,0,0)$ is in a span of the vectors I described. @Arturo Magidin: I'll fix this now. Sorry about this - I've forgotten a lot of linear algebra, and I tried to solve using a strange technique. –  Matt Groff Apr 2 '11 at 2:24

1 Answer 1

up vote 10 down vote accepted

The vectors $(a,a,a,0,0,0)$ are all multiples of $(1,1,1,0,0,0)$; the vectors $(b,b,b,b,b,b)$ are all multiples of $(1,1,1,1,1,1)$. The vectors $(0,d,2d,0,0,0)$ are all multiples of $(0,1,2,0,0,0)$. The span of all these vectors give you only a 3-dimensional subspace.

So the key lies in the vectors $(1,c,c^2,1,c,c^2)$; all these vectors span at most a $3$-dimensional subspace, since they all lie in the subspace of vectors $(x_1,x_2,x_3,x_4,x_5,x_6)$ with $x_1=x_4$, $x_2=x_5$, and $x_3=x_6$, which is $3$-dimensional. But they include $(1,1,1,1,1,1)$ (obtained with $c=1$); so you will get at best a 5-dimensional subspace from taking all these vectors together with the previously considered one; you cannot get a all of $\mathbb{R}^6$.

In fact, you get exactly a $5$-dimensional subspace: the vectors $$\begin{align*} &(1,c,c^2,1,c,c^2)\\ &(1,k,k^2,1,k,k^2)\\ &(1,\ell,\ell^2,1,\ell,\ell^2) \end{align*}$$ are linearly independent if and only if the vectors $(1,c,c^2)$, $(1,d,d^2)$, and $(1,\ell,\ell^2)$ are linearly independent. This occurs if and only if $$\left|\begin{array}{ccc} 1 & c & c^2\\ 1 & d & d^2\\ 1 & \ell & \ell^2 \end{array}\right| = (d-c)(\ell-c)(\ell-d)$$ is nonzero (this is a Vandermonde matrix); so distinct values of $c$, $d$, and $\ell$ will give you three linearly independent vectors, which therefore span $$\mathbf{W} = \bigl\{(x_1,x_2,x_3,x_4,x_5,x_6)\in\mathbb{R}^5\mid x_1=x_4, x_2=x_5, x_3=x_6\bigr\}.$$ So your vectors span exactly a five-dimensional subspace of $\mathbb{R}^6$, and not all of $\mathbb{R}^6$

In fact, $(0,0,1,0,0,0)$ will be one of the vectors that does not lie in the span: if it lay in the span, then so would $(0,1,0,0,0,0) = (0,1,2,0,0,0)-2(0,0,1,0,0,0)$; hence also $(1,0,0,0,0,0)=(1,1,1,0,0,0)-(0,1,0,0,0,0)-(0,0,1,0,0,0)$. Since you can get any vector of the form $(x,y,z,x,y,z)$, this would also allow you to obtain the other three standard basis vectors, and you would have a span equal to the entire space, which is not the case.

So, no, you cannot obtain $(0,0,1,0,0,0)$ with the vectors described.

share|improve this answer
    
Thanks so much for the expanded explanation. I have another unrelated algorithm that still has a chance... –  Matt Groff Apr 2 '11 at 2:42

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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