Do these polynomials $x-1$, $ (x-1)^2$ and $(x-1)^3$ span $p_3$? I heard my Math prof say that if a set is linearly independent than it is known to automatically span the vector set.
I expanded the eqn and wrote in the form of 
$$x^3 (c_3 ) + x^2 (c_2 - 3c_3 ) + x (c_1-2c_2+3c_3 ) - (c_1 + c_3 ) = 0$$
For this equation to turn to zero $c_1 = c_2 = c_3$  must be zero.
So it should be linearly independent and thus span but the answer is quiet opposite
Thanks!
 A: The error is in thinking that $P_3$ is of dimension $3$. Actually it is of dimension $4$ so independence of these does not imply that they span $P_3$. All three polynomials vanish at $x=1$ so any linear combination of them also vanishes at $1$. Hence they cannot span $P_3$. Any polynomial which does not vanish at $1$ (e.g. $1,\,x$ etc) cannot be written as a linear combination of these three. 
A: Those polynomial does not span $P_3$ (set of polynomials of the degree $\leq 3$) since you can not express constant polynomials with them. Say you can, then for each constant $k$ there are numbers $a,b,c$ such that $$k = a(x-1)+b(x-1)^2+c(x-1)^3$$
so $$ k = cx^3+(-3c+b)x^2+(a-2b+c)x+(-a+b-c)$$
But then $c=0$ and since $-3c+b=0$ also $b=0$ and since $a-2b+c=0$ also $a=0$, so $k=-a+b-c=0$.
So only the constant $0$ can be expressed with those polynomials.
A: To span $P_3$, any polynomial in $P_3$ must be a linear combination of the given polynomials:
$$ax^3+bx^2+cx+d=c_1(x-1)+c_2(x-1)^2+c_3(x-1)^3=\\
c_3x^3+(c_2-3c_3)x^2+(c_1-2c_2+3c_3)x+(-c_1+c_2-c_3) \Rightarrow \\
\begin{cases}a=c_3\\ b=c_2-3c_3 \\ c=c_1-2c_2+3c_3\\ d=-c_1+c_2-c_3\end{cases} \Rightarrow \begin{cases}c_3=a\\ c_2=b+3a \\ c_1=c+2(b+3a)-3a\\ d=-c_1+c_2-c_3=-(a+b+c)\end{cases},$$
which is a contradiction as it misses the cases $d\ne -(a+b+c)$.
