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Consider the basis {${u_1, u_2, u_3}$} of $\mathbb{R}^3$, where $u_1=(1,0,0)$, $u_2=(1,1,0)$ & $u_3=(1,1,1)$.Let {${f_1,f_2,f_3}$} be the dual basis of {${u_1, u_2, u_3}$} and $f$ be the linear functional defined by $f(a,b,c)=a+b+c$, $(a,b,c)\in \mathbb{R}^3.$ If $f=\alpha_1 f_1 + \alpha_2 f_2 + \alpha_3 f_3$, then find $(\alpha_1, \alpha_2, \alpha_3)$.

I worked out the dual basis here, which I got {${f_1,f_2,f_3}$}, where $f_1=(1,-1,0)$, $f_2=(0,1,-1)$ & $f_3=(0,0,1)$. Now, how should I proceed further to find $\alpha_1, \alpha_2, \alpha_3$ ?

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You've already donde all the hard work. What you've found is that for any $(x,y,z)\in\mathbb{R}^3$, \begin{align} f_1(x,y,z) &= x-y \\ f_2(x,y,z) &= y-z \\ f_3(x,y,z) &= z. \\ \end{align} Then, you need to find scalars such that \begin{align} f(x,y,z)=x+y+z &= \alpha_1f_1(x,y,z) + \alpha_2f_2(x,y,z)+ \alpha_3f_3(x,y,z)\\ &=\alpha_1(x-y) + \alpha_2(y-z)+ \alpha_3z\\ &= \alpha_1x + (\alpha_2-\alpha_1)y + (\alpha_3-\alpha_2)z. \end{align} which only solution is $\alpha_1=1$, $\alpha_2=2$ and $\alpha_3=3$.

You can also work the problem using your notation. Observe that $f=(1,1,1)$. So you want to find scalars such that \begin{align} f = (1,1,1) &= \alpha_1f_1+ \alpha_2f_2+ \alpha_3f_3\\ &=\alpha_1(1,-1,0) + \alpha_2(0,1,-1)+ \alpha_3(0,0,1)\\ &= \alpha_1(1,0,0) + (\alpha_2-\alpha_1)(0,1,0) + (\alpha_3-\alpha_2)(0,0,1). \end{align} Which clearly yields the same result.

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There is no work required at all.

Put ${\mathbb R}^3=:V$.

It is the very essence of the dual basis $(f_1,f_2,f_3)$ that the functionals $f_i\in V^*$ compute the coordinates $x_i$ of vectors $x\in V$ with respect to the chosen basis $(u_1,u_2,u_3)$: When $x=x_1u_1+x_2u_2+x_3 u_3$ then $x_i=\langle f_i,x\rangle$.

Now the basis $(u_1,u_2,u_3)$ of $V=(V^*)^*$ is in its own right the dual basis of $(f_1,f_2,f_3)$. Therefore the $u_j\,$, considered as functionals on $V^*$, compute the coordinates $\alpha_j$ of functionals $f\in V^*$ with respect to the basis $(f_1,f_2,f_3)$. Therefore we obtain $$\alpha_1=\langle f,u_1\rangle =a,\quad \alpha_2=\langle f,u_2\rangle =a+b,\quad \alpha_3=\langle f,u_3\rangle =a+b+c\ .$$

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