# Trying to find a basis for the span of 3 vectors

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!

• If two of them are linearly independent, they're a basis Jan 5, 2021 at 15:05
• 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. Jan 5, 2021 at 15:11

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 $$\text{Span}(\mathcal B)=\text{Span}(\{v_1,v_2,v_3\})$$.
in fact, $$(2,3,5)=3(1,1,2)-(1,0,1)$$.