Answer check (Vector space)

**scorpion007** · April 4th 2009, 01:11 AM

Hi, can someone just check my answer, please?

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.

**mr fantastic** · April 4th 2009, 03:32 AM

Originally Posted by scorpion007

Hi, can someone just check my answer, please?

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.

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