Basis for this $\mathbb{P}_3$ subspace. Just had an exam where the last question was:
Find a basis for the subset of $\mathbb{P}_3$ where $p(1) = 0$ for all $p$.
I answered $\{t,t^2-1,t^3-1\}$, but I'm not entirely confident in the answer.  Did I think about the question in the wrong way?
 A: Another way to get to the answer: 
$P_3=\{{ax^3+bx^2+cx+d:a,b,c,d{\rm\ in\ }{\bf R}\}}$. For $p(x)=ax^3+bx^2+cx+d$ in $P_3$, $p(1)=0$ is $$a+b+c+d=0$$ So, you have a "system" of one linear equation in 4 unknowns. Presumably, you have learned how to find a basis for the vector space of all solutions to such a system, or, to put it another way, a basis for the nullspace of the matrix $$\pmatrix{1&1&1&1\cr}$$ One such basis is $$\{{(1,-1,0,0),(1,0,-1,0),(1,0,0,-1)\}}$$ which corresponds to the answer $$\{{x^3-x^2,x^3-x,x^3-1\}}$$ one of an infinity of correct answers to the question. 
A: I'm assuming by $\mathbb{P}_3$ you mean the vector space of polynomials of degree 3 or less. It has dimension $4$, and since you have one condition (namely $p(1)=0$), we expect the subspace to have dimension 3, one less.
So it is enough to find three linearly independent basis vectors. In this example, that means three linearly independent polynomials, all with $p(1)=0$.
In your answer ($\{t, t^2-1, t^3-1\}$), the first one must be wrong, since $t(1)=1$. If you replace $t$ by $t-1$, then you have three linearly independent polynomials, all satisfying $p(1)=0$.
