Dimension of space spanned by row vectors

Question is to find dimension of spaces spanned by vectors :

$$\alpha_1=(1,1,0,1,0,0),\\ \alpha_2=(1,1,0,0,1,0),\\ \alpha_3=(1,1,0,0,0,1),\\ \alpha_4=(1,0,1,1,0,0),\\ \alpha_5=(1,0,1,0,1,0),\\ \alpha_6=(1,0,1,0,0,1).$$

I tried to make it down to row echelon form but it is not giving clear result... If i am not able to make a row zero that does not mean it can not be done... So, I am unale to conclude anything...

All I can see is that space should be of dimension at least four and it can not be six.

Please give hints to see this in less mechanical way.

Thank you.

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We see this as subspace of $\mathbb{R}^6$.. I was expecting that to be natural....I should have been mentioned it anyhow... – Praphulla Koushik Dec 23 '13 at 11:24
How do you expect it to be natural when you had a $\mathbb R^7$ vector in there? – Git Gud Dec 23 '13 at 11:26
Yes yes.. that was a typo.. i have corrected it – Praphulla Koushik Dec 23 '13 at 11:27
@B.S. : I am not sure about the argument.... – Praphulla Koushik Dec 23 '13 at 11:27
@CarstenSchultz could you make it as an answer? – Praphulla Koushik Dec 23 '13 at 11:35

Consider the matrix whose $i^\text{th}$ line is $\alpha _i$:

$$\begin{bmatrix} 1 & 1 & 0 & 1 & 0 & 0\\ 1 & 1 & 0 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 & 0 & 1\\ 1 & 0 & 1 & 1 & 0 & 0\\ 1& 0 & 1 & 0 & 1 & 0\\ 1 & 0 & 1 & 0 & 0 & 1\end{bmatrix}$$

Now look at it from the columns point of view.
Clearly the last three columns are linearly indepedent (as they are orthogonal).
Considering the fourth column together with to the last three, it can easily be seen that the set is still linearly indepedent.
Finally, the first column is a linear combination of the last three and the second one is a linear combination of the last four.
Therefore the vector space spanned by $\{\alpha _i\colon i\in\mathbb N \land 1\leq 7\}$ has dimension $4$.

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I think this satisfies the more natural approach... See the linear dependence of columns 2,3 and 4,5,6... +1 – Eleven-Eleven Dec 23 '13 at 11:49
This is perfect I would say :) This is so helpful... – Praphulla Koushik Dec 23 '13 at 11:50
@PraphullaKoushik Thank you. May I suggest, however, that you don't accept my answer right away? By accepting now, people who might have alternative methods might not want to share them and if (and when) they do post an answer, you might even like it better. This goes for all questions, no need to rush-accept. – Git Gud Dec 23 '13 at 11:52
@GitGud : Very valid point :) – Praphulla Koushik Dec 23 '13 at 11:53

I get the following row echelon form:

1  1 0  1  0 0
0 -1 1  0  0 0
0  0 0 -1  1 0
0  0 0  0 -1 1
0  0 0  0  0 0
0  0 0  0  0 0


Hence, I am not sure what your problem is.

Let me show you more mechanical stuff and how to use it to obtain more information about the space spanned by the vectors. First, from the row echelon form we can continue and after more elementary row operations obtain

1 0  1 0 0  1
0 1 -1 0 0  0
0 0  0 1 0 -1
0 0  0 0 1 -1
0 0  0 0 0  0
0 0  0 0 0  0


Now the rows of this matrix obviously span the solution space of the system of linear equations

\begin{align*} x_3&=x_1-x_2,\\ x_6&=x_1-x_4-x_5. \end{align*} (Make sure you see how to obtain the second equation.) The same holds for your original vectors. It is easy to check at least that they satisfy these equations and hence span a space of dimension at most $4$.

Also, if we start with the matrix whose columns are your vectors

1 1 1 1 1 1
1 1 1 0 0 0
0 0 0 1 1 1
1 0 0 1 0 0
0 1 0 0 1 0
0 0 1 0 0 1


and transform that to row echelon form

1  1  1  1  1  1
0 -1 -1  0 -1 -1
0  0 -1  0  0 -1
0  0  0 -1 -1 -1
0  0  0  0  0  0
0  0  0  0  0  0


then we see that the first four columns are independent, hence your $\alpha_1,\alpha_2,\alpha_3,\alpha_4$ are also independent. Also the last two columns are linear combinations of the first four. We can see these more clearly if we continue to

1 0 0 0 -1 -1
0 1 0 0  1  0
0 0 1 0  0  1
0 0 0 1  1  1
0 0 0 0  0  0
0 0 0 0  0  0.


Here we can read off that $\alpha_5=-\alpha_1+\alpha_2+\alpha_4$, $\alpha_6=-\alpha_1+\alpha_3+\alpha_4$. Again, easy to check.

Now the only thing that is not easy to check independently is that the first four vectors are indeed linearly independent. But at least we know from the first calculation that this must remain true if we delete the entries $x_3$ and $x_6$. So we would be left with the matrix

1 1 1 1
1 1 1 0
1 0 0 1
0 1 0 0


which is indeed regular.

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From what I gather the OP sees that as a process which too mechanical, he wants something more intuitive, a way that he can look at it and immediately guess what the dimension is. – Git Gud Dec 23 '13 at 11:45
@Carsten : This is perfectly alright But i was looking for a less mechanical way If i get a larger matric it would not be better to go for row echelon.... Thank you any how :) – Praphulla Koushik Dec 23 '13 at 11:48
Actually, if you have a large matrix with no apparent structure, then the mechanical way is the preferred one, since just looking at the matrix will not help. – Carsten S Dec 23 '13 at 11:50

The vectors $\alpha_1,\alpha_2,\alpha_3,\alpha_4$ are linearly independent. Also, $$\delta=\alpha_4-\alpha_1=(0,-1,+1,0,0,0)$$ satisfies $$\alpha_5=\alpha_2+\delta \text{ and }\alpha_6=\alpha_3+\delta.$$ Thus $\mathrm{Vect}(\alpha_1,\alpha_2,\alpha_3,\alpha_4)=\mathrm{Vect}(\alpha_1,\alpha_2,\alpha_3,\alpha_4,\alpha_5,\alpha_6)$ and $(\alpha_1,\alpha_2,\alpha_3,\alpha_4)$ is a basis. The space has dimension $4$.

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This does makes perfect sense... But, could this be made a bit more natural... – Praphulla Koushik Dec 23 '13 at 11:38
As Carsten showed below you end up with two rows of zeros, hence the dimension of the Span of your vectors is 4... What else do you need? – Eleven-Eleven Dec 23 '13 at 11:46
@ChristopherErnst : I was looking for a possibly non row echelon method result.... – Praphulla Koushik Dec 23 '13 at 11:48
See Gitgud's answer.... You can see the linear dependence immediately. – Eleven-Eleven Dec 23 '13 at 11:51