# Given functionals $q_1,q_2$ and $q_3$ in $\mathbb{R}^3$ determine if they form a basis of the dual space.

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?

• It is correct. If you need to justify this, you'll need to work a little bit more, though. – Giuseppe Negro Oct 18 '18 at 15:37

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.