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For each of the following, either use the subspace test to show that the given subset, $W$, is a subspace of $V$ , or explain why the given subset is not a subspace of $V$.

(c) $V = P_2(\mathbb{R})$ (real polynomial functions of degree at most $2$) and $W = \{f \in V | f(5) = f(-2) = 0 \}$.

I know the subspace test usually is:

1, is the set closed under addition? Usually:

If $(x_1,x_2,x_3) \in W$ and $(y_1, y_2, y_3) \in W$ then show that $(x_1+y_1, x_2+y_2, x_3+y_3) \in W$

2, is the set closed under scalar multiplication? Usually:

If $(x_1,x_2,x_3) \in W$ and $ \lambda \in \mathbb{R}$ then show that $\lambda \cdot (x_1,x_2,x_3) \in W$

3, is the set non-empty? Usually sets contain $O_v$

Now, I have easily done this for q1, q2 which I have not posted but am confused how to do it for this set.

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up vote 2 down vote accepted

$W$ is a subspace of $V$, where addition is given by $$(f+g)(x) := f(x) + g(x)$$ and scalar multiplication is given by $$(\lambda f)(x) :=\lambda \cdot f(x)$$ The $0_V$ is the zero-function ($f\equiv 0$).
To test this, let $f,g \in W$ and $\lambda \in \mathbb R$:

  1. $(f+g)(-2) = f(-2) + g(-2) = 0+0 = 0, (f+g)(5) = \ldots = 0$ ($\checkmark$)
  2. $(\lambda f)(-2) = \lambda f(-2) = \lambda \cdot 0 = 0, (\lambda f)(5) = \ldots = 0$ ($\checkmark$)
  3. $(0_V)(-2) = 0, (0_V)(5) = 0$ ($\checkmark$)
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