1) $\left \{P(X)\in R\left [ X \right ] , P(0)=1 \right \}$ in R[X] over R
2) $\left \{P(X)\in R\left [ X \right ] , P(1)=1 \right \}$ in R[X] over R
Those are vector spaces of all polynomials with real coefficients at least that's what my teacher said (this notation is strange to me). I've got problem visualising this since I've only worked on standard vectors so far like ${(x,y,z), x+y = z}$ and this seems abstract to me. Is it correct to do it like that? (on example 1.)
- Checking if $0$ belongs to the space
Let's take such polynomial that $P(0)=1$ so for instance $x^3+1$ then $P(-1)=0$ so $0$ does belong to that space(?), - Checking if $-P(x)$ belongs to that space,
take the same polynomial then
$-P(x)=-x^3 -1$ then $-P(0)=-1$ so $P(0)=1$ so this condition is fulfilled as well(?),
I'm not goint any further for now as I'd like to know if I'm even getting this right
Note that for polynomials vector spaces all work as for the ordinary ones replacing the vectors by $1,x,x^2,...$.
In the polynomials vector spaces the zero vector is represented by the zero polynomial $$P(x)=0 \quad \forall x$$
Then both subset are not subspaces since they do not contain the "zero vector".
As an alternative, we can easily show it by addition argument. Indeed the first subset is defined by the condition $P(0)=1$ then the polynomials in it are:
$$P(x)=1+a_1x+a_2x^2+...$$
$$Q(x)=1+b_1x+b_2x^2+...$$
note that
$$R(x)=P(x)+Q(x)\implies R(0)=2$$
thus it is not a subspace.
For the second let consider $P(x)$ and $Q(x)$ such that $P(1)=1$ and $Q(1)=2$ then
$$R(x)=P(x)+Q(x)\implies R(1)=2$$
thus it is not a subspace.