What it means for linear functionals to be linearly independent? Suppose we have the cannonical basis in dimension $3$, I guess three linear functionals are:
$$f_1(x,y,z)=x\\f_2(x,y,z)=y\\f_3(x,y,z)=z$$
Are they linearly independent? I'm confused because if I try to check it, then I get:
$$\alpha_1 f_1(x,y,z) + \alpha_2 f_2(x,y,z) + \alpha_3 f_3(x,y,z)=0 $$
Which yields:
$$\alpha_1 x + \alpha_2 y + \alpha_3z=0$$
It seems unlikely that for all $x,y,z$, the only solution is $\alpha_1=\alpha_2=\alpha_3=0 $ so either those are not LI or I am missing something very important. 
 A: The definition of linear (in)dependence is the same as it always is: for $f_1,f_2,$ and $f_3$ to be linearly dependent, there need to be scalars $\alpha_1,\alpha_2,\alpha_3$ (not all $0$) such that $$\alpha_1f_1+\alpha_2f_2+\alpha_3f_3=0.$$  What does this equation mean?  Well, it is an equation which takes place inside the vector space of linear functionals, so it is an equation of functions: it's saying the function $\alpha_1f_1+\alpha_2f_2+\alpha_3f_3$ is equal to the function that always outputs $0$.  Two functions are equal if they are the same on all inputs, so for this equation to hold, $$\alpha_1 f_1(x,y,z) + \alpha_2 f_2(x,y,z) + \alpha_3 f_3(x,y,z)=0$$ needs to be true for all $x,y,z$ and so $$\alpha_1 x + \alpha_2 y + \alpha_3z=0$$ needs to be true for all $x,y,z$.  But, for instance, if you set $x=1$ and $y=z=0$, this tells you that $\alpha_1=0$.  Similarly by other choices of $x,y,z$ you can conclude that $\alpha_2=0$ and $\alpha_3=0$ as well.  So no such $\alpha_1,\alpha_2,\alpha_3$ exist (that are not all $0$) and the functionals are linearly independent.
