Linearly independent subset? If $u,v,w$ and $z$ are distinct elements in $R$, then
 {$(1,u,u^2,u^3), (1,v,v^2,v^3), (1,w,w^2,w^3), (1,z,z^2,z^3)$} is a linearly independent subset of $R^4$.
Is that true or false? 
I even can't start the first step..
 A: If you haven't seen the Vandermonde matrix, then this exercise is a bit tricky. A usual strategy is to consider the 'same' problem, but in lower dimension. That is, instead of consider four vectors and ponder about independence in $\mathbb R^4$, consider constructing just three vectors, $(1,u,u^2),(1,v,v^2), (1,w,w^2)$, and ponder about independence in $\mathbb R^3$. Better yet, consider just two vectors, $(1,u)$ and $(1,v)$ first. Are these independent in $\mathbb R^2$? Yes, assuming that $u\ne v$. This is clear enough, but let's look at the determinant of the matrix with these two vectors as rows. It is easily seen to be equal to $v-u$. This shows again that the vectors are independent (since the determinant is non-zero). 
Now move on to consider the three vectors as above. Form the corresponding matrix and compute its determinant. While not immediate it's also not too difficult and you'll get a nice form for the determinant that will show easily it is non-zero. This will give you a hint as to what will happen with four vectors. The computation gets a bit messier but still doable by brute force. 
A: If they are linearly dependent, then the equation
$$
\underbrace{\pmatrix{1&u&u^2&u^3\\ 1&v&v^2&v^3\\ 1&w&w^2&w^3\\ 1&z&z^2&z^3}}_{A}
\pmatrix{a\\ b\\ c\\ d}=\pmatrix{0\\ 0\\ 0\\ 0}
$$
has a nontrivial solution $(a,b,c,d)$, i.e. there is a nonzero real polynomial $a+bx+cx^2+dx^3$ of degree $\le3$ that has at least four distinct roots $u,v,w,z$. This is impossible.
As pointed out by the others, The matrix $A$ in the above displayed equation is called a Vandermonde matrix. As we have demostrated, showing that it is invertible is easy (and the above proof works for any field, not just $\mathbb{R}$). The tricky part, though, is to find its determinant or inverse.
