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Let $V=R^3$, and define ${f_1,f_2,f_3} \in V^*$ as follows: $f_1(x,y,z)=x-2y, f_2(x,y,z)=x+y+z$, prove that ${f_1,f_2,f_3}$ is a basis for $V^*$.

I heard that all I need is to prove that ${f_1,f_2,f_3}$ are linearly independent and I can skip through the span part because $dim(V)=dim(V^*)$. Can someone explain why?

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    $\begingroup$ Because the because you wrote in your last line, of course...If two linear spaces are isomorphic then they have the same dimension, and if that dimension is finite, say $\;n\;$ , then any linearly independent set of $\;n\;$ vectors is a basis of the sapce $\endgroup$ – DonAntonio Apr 8 '20 at 1:05
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Try to find a basis $v_1,v_2,v_3$ for $V$ for which $f_1,f_2,f_3$ is its dual basis. Then, to see the independence, suppose that the zero functional, $0$, can be written as $$a_1f_1 + a_2f_2 + a_3f_3 = 0$$ for some scalars $a_1,a_2$ and $a_3$. Evaluating at $v_1,v_2$ and $v_3$ we obtain that $$0 = a_1f_1(v_1) + a_2f_2(v_1) = a_3f_3(v_1) = a_1$$ $$0 = a_1f_1(v_2) + a_2f_2(v_2) = a_3f_3(v_2) = a_2$$ $$0 = a_1f_1(v_3) + a_2f_2(v_3) = a_3f_3(v_3) = a_3$$ that is, they are linearly independent. Of course, you may think how to find $v_1,v_2$ and $v_3$. Hint : every $x \in V$ can be written as $$x = f_1(x)v_1 + f_2(x)v_2 + f_3(x)v_3$$ since $f_1,f_2$ and $f_3$ are just the coordinate functions with respect to that basis.

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In the finite-dimensional setting, there is a one-to-one correspondence between linear transformations and matrix representations. Hence, we can compute the dimension of the dual space, as follows: $$dim(V^{*}) = dim(\mathcal{L}(V,F)) = dim(V) \cdot dim(F) = dim(V).$$

In this problem, $V=\mathbb{R}^{3},$ so $dim(V^{*}) = 3.$

If we find a linearly independent subset of $V^{*}$ that contains exactly $3$ vectors, then it is a basis for $V^{*}.$ As discussed in the earlier solution by azif00, $\beta^{*} = \{f_{1},f_{2},f_{3}\}$ is a linearly independent subset of $V^{*}.$

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