Basis of span in $R^4$

Question

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?

Answer 1

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

Answer 2

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