Let $W=$Span$\{1, x\}$, $W_1=$Span$\{x^2, x^3\}$ and $W_2=$Span$\{1+x+x^2+x^3, 1+x+x^2-x^3\}$, how to show $P_3(R)=W\oplus W_1$ and $P_3(R)=W\oplus W_2$?
$P_3(R)=W+ W_1$ because Span$\{1, x\}$+Span$\{x^2, x^3\}$=$a+bx+cx^2+dx^3$, which represents any polynomial.
For $W\cap W_1=\{0\}$, I don't know how to show it.
$P_3(R)=W+ W_2$ because Span$\{1, x\}$+Span$\{1+x+x^2+x^3, 1+x+x^2-x^3\}$=$(a+c+d)+(b+c+d)x+(c+d)x^2+(c-d)x^3$ which represents any polynomial.
For $W\cap W_2=\{0\}$, again, I don't know how to show it.
Could someone fix my procedure?
$W\cap W_1=\{0\}$
To prove it, you need to find a non-null linear combination of $\{1, x\}$ and show you cannot get any non-null polynomial of $Span\{x^2, x^3\}$.
Indeed:
$a+bx$ can never be equal to $cx^2+dx^3$
$W\cap W_2=\{0\}$
To prove it, you need to find a non-null linear combination of $\{1, x\}$ and show you cannot get any non-null polynomial of $Span\{1+x+x^2+x^3, 1+x+x^2-x^3\}$.
Indeed:
$a+bx$ can never be equal to $c(1+x+x^2+x^3)+d(1+x+x^2-x^3)=(c+d)+(c+d)x+(c+d)x^2+(c-d)x^3$