1. ## Answer check (Vector space)

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

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

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

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

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

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

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

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

2. Originally Posted by scorpion007

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

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

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

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

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

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

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

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