<s>[INST] Find the dimension of the subspace of $\mathbb{R}^4$ spanned by $x_1= (1, 2, -1, 0)$, $x_2=(2, 5, -3, 2)$, $x_3=(2, 4, -2, 0)$, $x_4=(3, 8, -5, 4)$ 
Find the dimension of the subspace of $\mathbb{R}^4$ spanned by the vectors
  $$x_1= \begin{bmatrix}
1 & 2 & -1&  0  
      \end{bmatrix}^T, \qquad
x_2= \begin{bmatrix}
2 & 5 & -3&  2  
      \end{bmatrix}^T$$
$$x_3= \begin{bmatrix}
2 & 4 & -2&  0  
      \end{bmatrix}^T, \quad \text{and} \quad x_4= \begin{bmatrix}
3 & 8 & -5&  4  
      \end{bmatrix}^T.$$

The answer
The subspace $$Span(x_1,x_2,x_3,x_4)$$ is the same as the column space of the matrix
$$
X = \begin{pmatrix} 1 & 2 & 2 & 3 \\
2 & 5 & 4 & 8 \\
-1 & -3 & -2 & -5 \\
0 & 2 & 0 & 4
\end{pmatrix}
$$
The row echelon form of X is
$$
\begin{pmatrix}
1 & 2 & 2 & 3 \\
0 & 1 & 0 & 2 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}
$$
The first two columns x1,x2 of X will form a basis for the column space of X. Thus, dim $$Span(x_1 , x_2 ,x_3 ,x_4)=2$$
My question
I know how to reduce the row echelon form of X, but I have no idea how to know which coloums form a basis a column space and why not use row space?
 [/INST] The given vectors are the columns of
$$
A = 
\left[\begin{array}{rrrr}
1 & 2 & 2 & 3 \\
2 & 5 & 4 & 8 \\
-1 & -3 & -2 & -5 \\
0 & 2 & 0 & 4
\end{array}\right]
$$
The reduced row echelon form of $A$ is
$$
\operatorname{rref}(A)=\left[\begin{array}{rrrr}
1 & 0 & 2 & -1 \\
0 & 1 & 0 & 2 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array}\right]
$$
The first two columns of $\operatorname{rref}(A)$ are the pivot columns. This means that the first two columns of $A$ form a basis of $\operatorname{Col}(A)$.
Additionally, the nonzero rows of $\operatorname{rref}(A^\top)$ form a basis of $\operatorname{Col}(A)$. 
</s>