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.
 A: 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 $4$.
A: 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 $4$.
A: 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.
