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Let $v_1=(1,0,1)$, $v_2=(1,1,2)$, and $v_3=(2,3,5)$. I'm trying to find a basis for $span\{v_1,v_2,v_3\}$. I note that these vectors are not linearly independent because when I put them into a matrix the determinant is zero. What should I do from here? Thank you!

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    $\begingroup$ If two of them are linearly independent, they're a basis $\endgroup$ Jan 5, 2021 at 15:05
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    $\begingroup$ If you have vectors $v_1,v_2,v_3,\dots,v_n$ and you want to find a basis for the span of these vectors, we start building a collection of vectors... what I'll notate as $B$. We start with $B=\emptyset$. If $v_1$ is nonzero, then include $v_1$ in $B$, else don't. Next, if $B\cup \{v_2\}$ is still an independent set, then include $v_2$ into $B$... else don't. Continue this process for each $v_k$ in your original collection... if $B\cup \{v_k\}$ is still linearly independent, then include $v_k$, else don't and move on to the next... At the end of the day, what you have in $B$ will be a basis. $\endgroup$
    – JMoravitz
    Jan 5, 2021 at 15:11

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Note $c\cdot v_1=(c,0,c)\neq(1,1,2)$ and thus $v_1,v_2$ are linearly independent because $v_2\notin\text{Span}(\{v_1\})$. Therefore $\mathcal B:=\{v_1,v_2\}$ is a Hamel basis for $\text{Span}(\{v_1,v_2,v_3\})$ because $$v_3=-v_1+3v_2\in\text{Span}(\mathcal B)$$ and thus $\text{Span}(\mathcal B)=\text{Span}(\{v_1,v_2,v_3\})$.

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You are correct that the three vectors are not linearly independent;

in fact, $(2,3,5)=3(1,1,2)-(1,0,1)$.

But if you take any two of them, you should find that one is not a multiple of the other,

so they are linearly independent, so they are a basis for the span.

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