1. ## Answer check (Vector space)

Question:
Two subspaces S and T of $\displaystyle P_2(\mathbb{R})$ are given by

$\displaystyle S = \lbrace f(x) ~| ~3f(0)+f'(0) = 0 \rbrace$
$\displaystyle T = \lbrace f(x) ~| ~f(1) = 0 \rbrace$.

Obtain a non-trivial quadratic $\displaystyle g = ax^2 + bx +c$ such that $\displaystyle g \in S\cap T$.

My reponse:
Let $\displaystyle g = ax^2 + bx +c \in P_2(\mathbb{R})$,
Then $\displaystyle g' = 2ax + b$.

For membership in S: $\displaystyle 3c + b=0 ~ (1)$.
For membership in T: $\displaystyle a+b+c=0 ~ (2)$.

From (1), $\displaystyle b=-3c$.
From (2), $\displaystyle a=-b-c$.

Let c = 1.
$\displaystyle \therefore b=-3$ and $\displaystyle a=3-1=2$.

So $\displaystyle g=2x^2-3x+1 \in S\cap T$ is one such non-trivial quadratic.

2. Originally Posted by scorpion007

Question:
Two subspaces S and T of $\displaystyle P_2(\mathbb{R})$ are given by

$\displaystyle S = \lbrace f(x) ~| ~3f(0)+f'(0) = 0 \rbrace$
$\displaystyle T = \lbrace f(x) ~| ~f(1) = 0 \rbrace$.

Obtain a non-trivial quadratic $\displaystyle g = ax^2 + bx +c$ such that $\displaystyle g \in S\cap T$.

My reponse:
Let $\displaystyle g = ax^2 + bx +c \in P_2(\mathbb{R})$,
Then $\displaystyle g' = 2ax + b$.

For membership in S: $\displaystyle 3c + b=0 ~ (1)$.
For membership in T: $\displaystyle a+b+c=0 ~ (2)$.

From (1), $\displaystyle b=-3c$.
From (2), $\displaystyle a=-b-c$.

Let c = 1.
$\displaystyle \therefore b=-3$ and $\displaystyle a=3-1=2$.

So $\displaystyle g=2x^2-3x+1 \in S\cap T$ is one such non-trivial quadratic.