Proving the linear independence or dependence of a polynomial Here is the specific question I'm stuck on:
Let $P := R[t]≤2$ the real vector space of all polynomial functions of degree at most $2$ and consider $V := \mathbb{R}^P$, the real vector space of all functions from $P$ to $\mathbb{R}$. Determine the linear independence or dependence of the following lists $(f_1, f_2, f_3)$ in $V$:
(a) for $p \in P$, let $f_1(p) := p(0), f_2(p) := p(1)$ and $f_3(p) := p(2);$
(b) for $p \in P$, let $f_1(p) := p(0), f_2(p) := \int_0^1 p(t) \,\text{d}t$ and $f_3(p) := \int_{-1}^1p(t) \,\text{d}t.$
For part (a), is it enough to use the case of a zero degree vector as an example of a case where the polynomials would be linearly dependent, or how can I prove that all polynomials with degree less than $2$ would be linearly dependent?
Also for part (b), would the fact that the degree of the antiderivative is higher than the original function affect the linear dependence? I'm not really sure where to go with this part of the question.
 A: By definition:
a) Suppose there are $a,b,c\in\mathbb{R}$ such that $af_1+bf_2+cf_3=\vartheta$, where $\vartheta$ is such that $\vartheta(p)=0$ for all $p\in R[t]$.
Then, for $p(t)=1$ we have $0=\vartheta(p)=af_1(p)+bf_2(p)+cf_3(p)=ap(0)+bp(1)+cp(2)=a+b+c$. In the same way, for $p(t)=t$ we have $0=b+2c$ and for $p(t)=t^2$ we have $0=b+4c$. Solving the system give us $a=b=c=0$. Then $f_1,f_2$ and $f_3$ are l.i.
b) As we did for (a), you now will solve the system $a+b+2c=0$, $\frac{b}{2}+c=0$ and $\frac{b}{3}+\frac{2c}{3}=0$. Since this system has infinite solutions, $f_1$, $f_2$ and $f_3$ are l.d.
A: It might make things easier for you to convert these polynomials and linear functionals into something more familiar. If you represent a polynomial as a column vector, then the functionals can be thought of as row vectors, and their action on the polynomials becomes simple matrix multiplication.  
Given $p=a+bt+ct^2$, which we represent as $(a,b,c)^T$, we have in part (a) $f_1[p]=a$, $f_2[p]=a+b+c$ and $f_3[p]=a+2b+4c$. So, these three functionals correspoond to the row vectors $(1,0,0)$, $(1,1,1)$ and $(1,2,4)$, respectively, which are easily seen to be linearly independent. Similarly, for part (b) you can find row vector representations of the two integral operators and test these vectors for linear independence.
A: Consider the linear map $\varphi\colon\mathbb{R}^P\to\mathbb{R}^3$ given by
$$
\varphi(f)=\begin{bmatrix}
f(1)\\f(t)\\f(t^2)
\end{bmatrix}
$$
where $f(1)$ means the function $f$ evaluated at the constant polynomial $1$, $f(t)$ the function $f$ evaluated at the polynomial $t$ and similarly for $f(t^2)$.
Then, for case (a),
$$
\varphi(f_1)=\begin{bmatrix}
f_1(1)\\f_1(t)\\f_1(t^2)
\end{bmatrix}
=\begin{bmatrix}
1\\0\\0
\end{bmatrix}
$$
Similarly,
$$
\varphi(f_2)
=\begin{bmatrix}
1\\1\\1
\end{bmatrix},
\qquad
\varphi(f_3)
=\begin{bmatrix}
1\\2\\4
\end{bmatrix}
$$
Since the matrix
$$
\begin{bmatrix}
1 & 1 & 1\\
0 & 1 & 2\\
0 & 1 & 4
\end{bmatrix}
$$
has rank $3$, we have that $(\varphi(f_1),\varphi(f_2),\varphi(f_3))$ is a linearly independent list in $\mathbb{R}^3$, so also $(f_1,f_2,f_3)$ is linearly independent.
For case (b), we have
$$
\varphi(f_1)
=\begin{bmatrix}
1\\0\\0
\end{bmatrix},
\qquad
\varphi(f_2)
=\begin{bmatrix}
1\\1/2\\1/3
\end{bmatrix},
\qquad
\varphi(f_3)
=\begin{bmatrix}
2\\0\\2/3
\end{bmatrix}
$$
and the matrix
$$
\begin{bmatrix}
1 & 1 & 2 \\
0 & 1/2 & 0 \\
0 & 1/3 & 2/3
\end{bmatrix}
$$
again has rank $3$.
