Meaning of definition of a linear subspace of $\mathrm{P}_2(x)$ Show that $W$ is a subspace of $\mathrm{P}_2(x)$, where $W = \{ p(x)\, ; \, p(0) = \alpha p(1) \}$ and $\alpha \in \mathbb{R}$. I know that I have to show that $W$ is closed under addition and scalar multiplication, but I'm not quite sure what $p(0) = \alpha p(1)$ means. Given that $\mathrm{P}_2(x)$ is all real coefficient polynomials  of order $\le 2$, what kind of polynomials subset is is $W$?

$$p(0) = \alpha p(1)$$
$$\iff c = \alpha(a + b + c)$$
$$\implies W = \big\{ p(x) = ax^2 + bx + \alpha(a + b + c) \big\} $$
To show a set is a subspace, you have to show the 3 following:


*

*W is non-empty


*W is closed under addition


*W is closed under scalar multiplication.

W is non-empty
You usually prove this by showing the $0$ vector of the superspace belongs to the subspace.
In fact,
$$0 = 0x^2 + 0x + 0$$ and $0 = \alpha(0 + 0+ 0)$
$\implies 0 \in W$
W is closed under scalar multiplication and addition
Let $p(x)$ and q(x) $\in$ $W$. They satisfy the property of $W$ namely,

*

*$p(x) = ax^2 + bx + c = ax^2 + bx + \alpha(a + b + c)$


*$q(x) = dx^2 + ex + f =dx^2 + ex + \alpha(d + e + f)$
Now
$$p(x) + kq(x) = ax^2 + bx + \alpha(a + b + c) + kdx^2 + kex + k\alpha(d + e + f)$$
$$= (a + kd)x^2 + (b + ke)x + \alpha(a + b + c+ kd + ke + kf)$$
Take
$$A = a + kd$$
$$B = b + ke$$
$$C = c + kf$$
$$\implies p(x) + kq(x) = Ax^2 + Bx + \alpha(A + B + C)$$
Thus $p(x) + kq(x) \in W$
You can now conclude W is a subspace of $P_2$
