91影视

A subsetof a linear space Vis called a subspace of Vif

1. contains the neutral element 0of V.
2. is closed under addition (if fand gare in Wthen so is f+g)
3. is closed under scalar multiplication (if fis in Wand kis scalar, then kfis in W).

we can summarize parts b and c by saying that W is closed under linear combinations.

General expression for a quadratic polynomial with one variable is$\mathit{p}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{a}{\mathit{t}}^{\mathbf{2}}\mathbf{+}\mathit{b}\mathit{t}\mathbf{+}\mathit{c}$ 鈥︹�� (1)

Given$p\left(2\right)=0.$ .

Substitute 0 forin equation (1),

$$\begin{gathered} p\left(2\right)=4a+2b+c \\ 0=4a+2b+c \\ c=-4a-2b \end{gathered}$$

Therefore, $p\left(t\right)=a{t}^2+bt-4a-2b.$.

We know, $p\left(t\right)=a{t}^2+bt-4a-2b.$.

Assume $q\left(t\right)=e{t}^2+dt-4e-2d.$.

Let us assume two scalars $伪,尾$,

Then, we can write,

$$\begin{gathered} 伪p\left(t\right)+尾q\left(t\right)=伪\left(a{t}^2+bt-4a-2b\right)+尾\left(e{t}^2+dt-4e-2d\right) \\ =伪a{t}^2+伪bt-伪4a-伪2b+尾e{t}^2+尾dt-尾4e-尾2d \\ =t^2\left(伪a+尾e\right)+t\left(伪b+尾d\right)-4\left(伪a+尾e\right)-2\left(伪a+尾d\right) \end{gathered}$$

Therefore, the given set is closed under linear combinations. Hence, it is a subspace of $P_2$.