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What's the method for finding basis where $\beta_1=(0,2,1) \ , \ \beta_2=(1,1,2)$ and $\beta_1$ has cooridnates $1,2,-1$ and $\beta_2$ has $0,0,1$ ?

we have that $(1,1,2)=(a_3,b_3,c_3)$ and $(0,2,1)=(a_1,b_1,c_1)+2(a_2,b_2,c_2)-(1,1,2)$ so

$(1,3,3)=(a_1+2a_2,b_1+2b_2,c_1+2c_2)$ and can I chose here random numbers for $a_1,a_2,a_3$ and calculate the rest ?

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  • $\begingroup$ yes, but you should choose them, so that you get linearly independent elements. $\endgroup$ – Kamal Saleh Nov 27 '14 at 20:57
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Well, $(a_3,b_3,c_3)=(1,1,2)$ so those three values are determined.

As you said, for the remaining six unknowns $a_i,b_i,c_i$, $i=1,2$, you have three equations, so the system is underdetermined. It is also completely decoupled.

\begin{align} a_1+2a_2&=1\\ b_1+2b_2&=3\\ c_1+2c_2&=3 \end{align}

Thus, $a_1=1-2a_2$, $b_1=3-2b_2$, and $c_1=3-2c_2$. Choose any values of $a_2,b_2,c_2$ such that the resulting $(a_i,b_i,c_i)$ is linearly independent and you have found a basis that meets your criteria.

For example, $a_2=1,b_2=0,c_2=-1$ yields the basis $\{(-1,3,5),(1,0,-1),(1,1,2)\}$.

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