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$$\text{let }v_1= \begin{bmatrix} 1 \\ 1 \\ -1 \\ 1 \end{bmatrix} , v_2 = \begin{bmatrix} -1 \\ 1 \\ 1 \\ -1 \end{bmatrix} v_3 = \begin{bmatrix} 1 \\ -3 \\ -1 \\ 2 \end{bmatrix} v_4 = \begin{bmatrix} 1 \\ -2 \\ -1 \\ 3 \end{bmatrix} v_5 = \begin{bmatrix} 1 \\ 2 \\ 0 \\ 3 \end{bmatrix} \text{ Find a subset of} $$

$v_1, v_2, v_3, v_4, v_5$ that form a basis for $sp(v_1, v_2, v_3, v_4, v_5)$

My solution:

What I did is I row reduced and got

$$ \begin{bmatrix} 1 & 0 & 0 & 3/2 & 0 \\ 0 & 1 & 0 & 5/2 & 0 \\ 0 & 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix} $$

My final answer is the basis is $\{v_1, v_2, v_3, v_5\}$

Because I think the basis is like the column space and there is a pivot in those columns. Am I right?

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Because I think the basis is like the column space and there is a pivot in those columns.

I'm not really sure what you mean by "is like the columns space", but your method is fine.

What I did is I row reduced and got

$$ \begin{bmatrix} 1 & 0 & 0 & \color{blue}{3/2} & 0 \\ 0 & 1 & 0 & \color{blue}{5/2} & 0 \\ 0 & 0 & 1 & \color{blue}{2} & 0 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix} $$

My final answer is the basis is $\{v_1, v_2, v_3, v_5\}$

That's correct. You can interpret the fourth column as follows: $$v_4 = \color{blue}{\tfrac{3}{2}}v_1+\color{blue}{\tfrac{5}{2}}v_2+\color{blue}{2}v_3$$

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