Studying for an exam and this question appeared on a previous exam.
Let $P_4$ be the space of all polynomials of degree at most 4. The subspace $V$ of $P_4$ consists of all polynomials $p(t)$ such that $p(0) = 2p(1)$. Dimension of $V$ is$\dots$
The answer is 4, but I'm not sure I understand. I fairly sure the dimension would be 5 if we were only considering $P_4$, but it seems $p(0) = 2p(1)$ is affecting it in some way. Any advice?
We have that
$$p(x)=ax^4+bx^3+cx^2+dx+e$$
and
$$p(0)=2p(1)\implies e=2a+2b+2c+2d+2e \implies e=-2a-2b-2c-2d$$
that is
$$p(x)=ax^4+bx^3+cx^2+dx-2a-2b-2c-2d=\\=a(x^4-2)+b(x^3-2)+c(x^2-2)+d(x-2)$$
and the basis is $(x^4-2),(x^3-2),(x^2-2),(x-2)$, therefore the dimension is $4$.