# Notes for Snow Zhou

**6** Let
$$\mathbb P_3(\mathbb R):=\{a+bx+cx^2+dx^3\}.$$
Let $W=\mathrm{span}\{1,x\}$ and $W_2=\mathrm{span}\{1+x+x^2+x^3,1+x+x^2-x^3\}$. Show that
$$
\mathbb P_3(\mathbb R)=W+W_2.
$$

**Proof**: By definition of the sum of subspaces, it suffices to show that every vector $f\in \mathbb P_3(\mathbb R)$ can be written as
$$
f(x)=w(x)+u(x)\qquad\qquad (*)
$$
for some vector $w\in W$ and $u\in W_2$. Indeed, since $f\in \mathbb P_3(\mathbb R)$, there exist coefficients $a,b,c,d\in\mathbb R$ such that
$$
f(x)=a+bx+cx^2+dx^3.
$$
In order to construct $w\in W$ and $u\in W_2$ satisfying equation (*), we simply take
\begin{equation}
\begin{split}
u(x):=\left(\frac{d+c}2\right)\cdot(1+x+x^2+x^3) + \left(\frac{c-d}2\right)\cdot(1+x+x^2-x^3)\\\in \mathrm{span}\{1+x+x^2+x^3,1+x+x^2-x^3\}=W_2,
\end{split}
\end{equation}
and
$$
w(x):= (a-c)\cdot 1+(b-c)\cdot x \in \mathrm{span}\{1,x\}=W.
$$
Then by calculation, we have
$$
w(x)+u(x)=f(x)
$$
as desired. Therefore $\mathbb P_3(\mathbb R)=W+W_2$.