# 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 $V^*$.

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?

• 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 Apr 8, 2020 at 1:05

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.
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 $$V^{*}.$$