Real polynomials forming a basis How do you show that a sequence of polynomials form a basis? I know that the rref criterion of a basis (no $0$ rows and has a leading 1 in every column) only works for $K^n$. Particularly,
Consider the set Consider the set $V$ of real polynomials $f$ of degree at most
3 which satisfy $f(1) = f(2)$, and $V$ is a vector space over $\mathbb{R}$.
Show that the following polynomials $f_1(x) = 1$, $f_2(x) = x^2-3x$
and $f_3(x) = x^3-7x$ form a basis of V .
Note: I can't use the determinant method to determine independence.
PROVING V is a vector space:
Let $f(1)=f(2)$, $g(1)=g(2)$ and $\alpha, \beta \in \mathbb{R}$
$(\alpha f+ \beta g)(1)$
= $(\alpha f)(1) + (\beta g)(1)$
= $\alpha (f(1)) + \beta (g(1))$
= $\alpha (f(2)) + \beta (g(2))$
= $(\alpha f)(2) + (\beta g)(2)$
= $(\alpha f + \beta g)(2)$
 A: You need to show:

*

*The dimension of $V$ is $3$

*$f_1, f_2$ and $f_3$ are in $V$

*$f_1, f_2$ and $f_3$ are linearly independent.
i.e. The only linear combination $a_1f_1 + a_2f_2 + a_3f_3=0$ is the trivial combination.

A: Assuming your vectors are in $V$. There are two conditions.

*

*Linear independence. Let $a,b,c\in\mathbb R$ be such that
$$\forall x\in\mathbb R,\quad a + b(x^2-3x) + c(x^3-7x) = 0. $$
Does it follow that $a=b=c=0$? Note that this equality holds for any $x$. Fix three different $x$ and get 3 equations.

*Spanning. Let $p(x) := ax^3 + bx^2+cx + d$ such that $p(1) = p(2)$. Find $u,v,w$ such that
$$\forall x\in\mathbb R,\quad ax^3+bx^2+cx + d = u + v(x^2-3x) + w(x^3-7x). $$

To check $V$ is a vector subspace, take $f,g\in V$ i.e $f(1) = f(2)$ and $g(1) = g(2)$. Check the condition for $V$:
$$(f+g)(1) \overset{?}= (f+g)(2),\qquad \forall \alpha\in\mathbb R,\quad (\alpha f)(1) = (\alpha f)(2)\,?  $$
or you could do this in one go by verifying
$$ \forall \alpha,\beta\in\mathbb R,\quad (\alpha f+\beta g)(1) = (\alpha f+\beta g)(2). $$
Either option suffices.
