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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.

Git Gud
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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... –  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 –  Dec 23 '13 at 11:27
  • @B.S. : I am not sure about the argument.... –  Dec 23 '13 at 11:27
  • @CarstenSchultz could you make it as an answer? –  Dec 23 '13 at 11:35
  • Please stop posting exam problems Prafhulla Koushik. I hate this practice. Please wait a few days to come the test results. I hope it will never help anybody including you, in next four months. – Supriyo Dec 23 '13 at 13:07
  • @HopelessFool I would not bother if you hate this practice.... This is just my personal concern and this has nothing to do with "Exam". I am just learning something ... –  Dec 23 '13 at 13:17
  • @HopelessFool : I am concerned about your feelings in fact anybody else's feelings... But i can not help in this case... –  Dec 23 '13 at 13:30
  • Not an unexpected answer. – Supriyo Dec 23 '13 at 16:20
  • Hope your successful "learning". My comments will not create any trouble. – Supriyo Dec 23 '13 at 16:21
  • @HopelessFool : Do not take my comments so seriously, it would give you nothing but pain... I totally understand your hate/frustration but that is not for which i have posted these questions.... Hope this will not effect our concern for one another... Hope you understand..... I do not even want to call you by "Hopeless fool" so i am sure i am not doing this intentionally... good luck! –  Dec 24 '13 at 01:31
  • I am glad because you have understood my feeling. I have made a mistake in this problem. I took 6 instead of 4. To get accurate answer writing the matrix is sufficient. There is no place to write it down. So mistake. When I had seen this question here and I knew my fault, I... you have understood. User name can not be changed before one month. So you shall call me by a new name from the next January. – Supriyo Dec 24 '13 at 01:53
  • Hope January comes a bit fast..... –  Dec 24 '13 at 01:56

4 Answers4

4

Consider the matrix whose $i^\text{th}$ row 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 $4$.

Git Gud
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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... –  Dec 23 '13 at 11:50
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    @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 :) –  Dec 23 '13 at 11:53
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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 $4$.

Olivier Bégassat
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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.

Carsten S
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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 :) –  Dec 23 '13 at 11:48
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    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
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When we write the given vectors in a matrix form and reduce it to a row echelon thus obtaining the rank as 4 where the 1 St four columns are linearly independent implies that the first four scalars are zero..so out of the 6 vectors given when only 4 are linearly independent..we can say that the vector space is spanned by linearly independent vectors alone that is the number 4.. so dim of the v.sp should be 4.

Anahita Garrad
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