I have the following linear algebra question on direct sums: I am given the vector spaces: $ V = R_4[x] $
$ W = span\{x^4-x^2,3x^4-x^3+1 \}$
I am asked to find the complement to the direct sum i.e. a vector space U spanned by subset of $\{1,x,x^2,x^3,x^4\}$ such that we have: $ V = U \oplus W $ Could you please assist me in providing the solution and the background of how you get there please? Thank you all
Hint:
Let $\mathbf{u}=x^4-x^2$ and $\mathbf{v}=3x^4-x^3+1$ Then $\mathbf{u}$ and $\mathbf{v}$ when combined linearly will either produce a $4^{\text{th}}$ degree polynomial or a $3^{\text{rd}}$ polynomial ($-3\mathbf{u}+\mathbf{v}$). This suggests that $U$ is possibly generated by $\{1,x,x^2\}$. Hope this helps you get started in the right direction.
Added explanation:
The idea of decomposing a vector space into direct sum is to try representing every vector $v \in V$ as a sum of two vectors $u+w$ such that $u \in U$ and $w \in W$ and $U \cap W = \{\vec{0}\}$. In your question I first looked at $W$ and tried to find the kind of polynomials which can live in $W$. Based on the degree of the polynomials in the spanning set I concluded that only degree $3$ and degree $4$ polynomials will live in the subspace $W$ (of course apart from $\vec{0} \in W$). So in order to get $U$ I need to ensure two things: except for the zero vector $U$ doesn;t share anything with $W$ and whatever things are leftover in $V$ (that are not covered by $W$) should come from $U$. I hope this helps clarify some of the doubts you might have had.