Finding a basis for all polynomials $f(t)$ of degree 3 such that $f(1)=0$ and $\int_{-1}^{1} f(t) dt$=0 this is a problem from my textbook I am working on. I am not sure how to approach it -- I would like to solve it using a 'matrix' way if possible -- that is, to set up a matrix with coefficients etc. and solve for the basis of the kernel, which would be the basis of what I am looking for (not sure if that's possible).
How to find a basis for all polynomials $f(t)$ of degree 3 such that $f(1)=0$ and $\int_{-1}^{1} f(t) dt$=0
Also, my book defines polynomials of degree n as subspaces/vector spaces, but Mariano in the following question argues that it is not such because it does not contain the zero element (though I think if we set all the coefficients to 0, it does), so now I am confused as to whether the problem even makes any sense:
https://math.stackexchange.com/questions/1350/are-the-polynomials-of-degree-n-a-vector-space
 A: Almost certainly, your textbook is refering to the set of all polynomials of degree at most $n$, rather than the polynomials of degree exactly $n$. The latter is indeed not a vector space under the usual addition of polynomials: it doesn't contain the $0$ polynomial (which either has no degree, or has some degree smaller than $0$), and it's not closed under sums: both $x^n+x^{n-1}$ and $-x^n$ are polynomials of degree $n$, but their sum is not.
So I suspect your textbook is asking for the vector space that consists of all polynomials of degree at most $3$, which further satisfy $f(1)=0$ and $\int_{-1}^1 f(t)\,dt = 0$. That is, if we let $\mathbf{P}_3$ denote the vector space of polynomials of degree at most $3$, then it wants you to find a basis for $\mathbf{W}$, where
$$\mathbf{W} = \left\{ f(t)\in\mathbf{P}_3\;\left|\; f(1)=0\text{ and }\int_{-1}^1f(t)\,dt = 0\right.\right\}.$$
You say you want to solve this via a matrix; well: the map $\mathbf{P}_3\to\mathbb{R}$ given by "evaluation at $1$" ($p(t)\longmapsto p(1)$) is a linear transformation. And the map $\mathbf{P}_3\to\mathbb{R}$ given by "integrate over $[-1,1]$ ($p(t)\longmapsto \int_{-1}^1p(t)\,dt$) is also a linear transformation. Call the first linear transformation $E$ and the second $T$. The polynomials that are in the nullspace/kernel of the first map are precisely the ones that satisfy $f(1) = 0$. The ones that are in the kernel of the second one are precisely the ones that satisfy $\int_{-1}^1f(t)\,dt = 0$. You want the ones that are in the kernel of both.
The standard way of achieving this is to map to the product: take the map $L\colon\mathbf{P_3}\to \mathbb{R}\times\mathbb{R}=\mathbb{R}^2$, given by:
$$L(p(t)) = \Bigl( E(p),\; T(p)\Bigr) = \left(p(1),\quad\int_{-1}^1p(t)\,dt\right).$$
Then the subspace you want is precisely the kernel of $L$. 
To find the kernel of $L$, one possibility is to pick a basis for $\mathbf{P}_3$ (say, the standard basis $[1,t,t^2,t^3]$), a basis for $\mathbb{R}^2$ (say, the standard basis $[(1,0), (0,1)]$), construct the matrix representation for $L$, and find the nullspace of that matrix. Interpreting this nullspace in terms of the basis $[1,t,t^2,t^3]$ will give you the kernel you want, and finding a basis for that nullspace will yield a basis for the subspace of $\mathbf{P}_3$ that you are looking at.
A: A basis of the linear subspace of polynomials $f$ of degree at most $3$ such that $f(1)=0$ is $(p_1,p_2,p_3)$ where
$$
p_1(t)=2(1-t),\quad
p_2(t)=3(1-t)^2,\quad
p_3(t)=4(1-t)^3.
$$
(Proof: Let $p_0(t)=1$. Then $(p_0,p_1,p_2,p_3)$ is a basis of the linear subspace of polynomials $f$ of degree at most $3$, as every family of polynomials with degree $0$, $1$, $2$ and $3$.)
Now, the integral of $a_1p_1+a_2p_2+a_3p_3$ over $[-1,1]$ is zero iff $4a_1+8a_2+8a_3=0$ iff $a_1=-2a_2-4a_3$, hence a basis of the subspace the OP is interested in is $(2p_1-p_2,4p_1-p_3)$, or equivalently, $(q_2,q_3)$ with
$$
q_2(t)=(1-t)(1+3t),\quad
q_3(t)=(1-t)(1+2t-t^2).
$$
