Find a basis for the subspace of $\mathbb R^4$ spanned by the 4 vectors. Find a basis for the subspace of $\mathbb R^4$ spanned by the 4 vectors.
$\alpha_1=(1,1,2,4),\alpha_2=(2,-1,-5,2),\alpha_3=(1,-1,-4,0),\alpha_4=(2,1,1,6)$
Check for linear dependence : $ c_1\alpha_1+c_2\alpha_2+c_3\alpha_3+c_4\alpha_4=0$
$\left[\begin{matrix}
1 & 2 & 1 & 2 & 0 \\ 1 & -1 & -1 & 1 & 0 \\ 2 & -5 & -4 & 1 & 0 \\ 4 & 2 & 0 & 6 & 0 \\ 
\end{matrix}\right]$ after row reduction became $\left[\begin{matrix}
1 & 0 & -1/3 & 4/3 & 0 \\ 0 & 1 & 2/3 & 1/3 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 
\end{matrix}\right]$
$\Rightarrow c_3,c_4$ can be any value, and $c_1=(1/3)c_3+(4/3)c_4, c_2=-(2/3)c_3-(1/3)c_4$, the given vectors are dependent.  
But how to find basis?
$\Box\left[\begin{matrix}1/3 \\ -2/3 \\ 1 \\ 0\end{matrix}\right]+\Box \left[\begin{matrix}-4/3\\-1/3\\0\\1\end{matrix}\right]=\alpha_k, k=1,2,3,4$ is not helping
 A: Operate row reduction on the transposed matrix, i.e. write the vectors as row vectors:
$$\begin{bmatrix}
1&1&2&4\\
2&-1&-5&2\\
1&-1&-4&0\\
2&1&1&6
\end{bmatrix}\rightsquigarrow
\begin{bmatrix}
1&1&2&4\\
0&-3&-9&-6\\
0&-2&-6&-4\\
0&-1&-3&-2
\end{bmatrix}\rightsquigarrow
\begin{bmatrix}
1&1&2&4\\
0& 1&3&2\\
0&-2&-6&-4\\
0&-1&-3&-2
\end{bmatrix}
\rightsquigarrow\begin{bmatrix}
1&1&2&4\\
0& 1&3&2\\
0&0&0&0\\
0&0&0&0
\end{bmatrix}$$
Thus the vectors $\;\alpha_1=(1,1,2,4)$ and $\;\beta_2=(0,1,3,2)$ constitute a basis of the subspace spanned by $\;\alpha_1, \alpha_2, \alpha_3, \alpha_4$.
As $\beta_2=\alpha_1-2\alpha_2$, we might as well take $(\alpha_1, \alpha_2)$ as a basis. Or, proceeding a little further in row reduction, set $\;\beta_1=\alpha_1-\beta_2=(1,0,-1,2)$ and choose $\;(\beta_1,\beta_2)$ as a basis.
A: As you have shown, the dimension of the linear subspace $V$ created by the vectors $\left<\alpha_1,\alpha_2,\alpha_3,\alpha_4\right>$ is equal to 2 (since $\text{rank}(A)=2$, where $A=(\alpha_1,\alpha_2,\alpha_3,\alpha_4)$ is the matrix created by the fouw vectors as columns and rank of a matrix is the minimum number of linearly independent lines or columns). So, we have to find two lineraly independent vectors that can generate $V$, since $\dim_{\mathbb{R}}V=2$.
We see that $\alpha_2\neq\lambda\alpha_1$ for every $\lambda\in\mathbb{R}$, so $\alpha_1$ and $\alpha_2$ are linearly independent. Therefore, a basis for $V$ must be $\mathbf{v}=\{\alpha_1,\alpha_2\}$. 
Note that $\mathbf{v}$ is really a base since we can write $\alpha_3$ and $\alpha_4$ as linear combinations of elements of $\mathbf{v}$, as follows:


*

*$\alpha_3=-\frac{1}{3}\alpha_1+\frac{2}{3}\alpha_2$,

*$\alpha_4=\frac{4}{3}\alpha_1+\frac{1}{3}\alpha_2$

