Basis of span in $R^4$ I have a set of 3 vectors in $\mathbb{R}^4$
$$v_1 = (1, 0, 1, 0)^T, v_2 = (2, h, 2, h)^T, v_3 = (1, 1+h, 1, 2h)^T$$
and I am to find the basis of the span of $v_1, v_2, v_3$
So far I have set it up in a matrix in the form of
$$\begin{bmatrix} 
1 & 2 & 1 \\
0 & h & 1 + h \\ 
1 & 2 & 1 \\
0 & h & 2h
\end{bmatrix}
$$
and through Gaussian elimination I have come to the result 
$$\begin{bmatrix} 
1 & 2 & 1 \\
0 & h & 1 + h \\ 
0 & 0 & h - 1 \\
0 & 0 & 0
\end{bmatrix}
$$
Leading me to think that the Basis for this span is the set of the vectors $v_1$ and $v_2$ because of the linear dependence. Have I made some mistake here or completely misunderstood what the basis is?
 A: You have to consider the following cases


*

*for $h=1$ the first two columns vectors are liner independent thus $v_1$ and $v_2$ are a basis

*for $h=0$ the first and the third columns vectors are liner independent thus  $v_1$ and $v_3$ are a basis

*otherwise the first three columns vectors are liner independent thus $v_1$,$v_2$ and $v_3$ are a basis

A: It looks like you (also) made row operations, which are not allowed here in the following sense:
For example $e_1=\begin{bmatrix}1\\0\end{bmatrix}$ is a basis of $\Bbb{R}e_1$ but $\begin{bmatrix}1\\1\end{bmatrix}$ isn't a basis of $\Bbb{R}e_1$. 
I would do the following:
$$\begin{bmatrix} 
1 & 2 & 1 \\
0 & h & 1 + h \\ 
1 & 2 & 1 \\
0 & h & 2h
\end{bmatrix}
\rightarrow\begin{bmatrix} 
1 & 0 & 0 \\
0 & h & 1 + h \\ 
1 & 0 & 0 \\
0 & h & 2h
\end{bmatrix}
\rightarrow\begin{bmatrix} 
1 & 0 & 0 \\
0 & h & 1 - h \\ 
1 & 0 & 0 \\
0 & h & 0
\end{bmatrix}
$$
So you have the following three cases


*

*$h=0$: The first and the third column form a basis

*$h=1$: The first and the second column form a basis

*Else: The three columns form a basis

