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If I have the following linear functionals: $$q_1(x,y,z)=2x-y+3z$$ $$q_2(x,y,z)=3x-5y+z$$ $$q_3(x,y,z)=4x-7y+z$$ then in order to show that the linear functionals form a basis can I just find the determinant of the matrix $$\left[\begin{matrix} 2 & -1 & 3\\ 3 & -5 & 1 \\ 4 & -7 & 1 \end{matrix} \right]$$ of this matrix and show that it is not equal to zero. Will this solution work?

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    $\begingroup$ It is correct. If you need to justify this, you'll need to work a little bit more, though. $\endgroup$ – Giuseppe Negro Oct 18 '18 at 15:37
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Let $\alpha:=\{(1,0,0),(0,1,0),(0,0,1)\}=\{e_1,e_2,e_3\}$ denote the standard basis of $\mathbb R^3$. Then any linear function $f$ from $\def\rt{\mathbb R^3}\rt$ to $\mathbb R$ can be uniquely written as $f(e_1)f_1+f(e_2)f_2+f(e_3)f_3$, where $f_i$ is the function defined as $f_1(x,y,z)=x,\quad f_2(x,y,z)=y,\quad f_3(x,y,z)=z$. In fact, $\bar\alpha:=\{f_1,f_2,f_3\}$ is the dual basis corresponding to $\alpha$.

Then the matrix in question is the matrix representation of the functions $q_i$ with respect to this basis $\bar\alpha$. We know that these three elements form a basis if and only if the matrix of them with respect to another basis is invertible, hence if and only if the matrix under discussion has non-zero determinant. Therefore your solution works.


Hope this clarifies some doubts.

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