$S = \{p\in p_3; p(x) = a_0 + a_1x^2 +a_3x^3\}$
Let
$$p(x) = a_0 + a_1x^2 +a_3x^3 \in S\\q(x) = b_0 + b_1x^2 +b_3x^3 \in S$$
and $p(7) = 0 \in S$
-In order to show that this is a subspace I just have to show that $p(x) + q(x) \in S$ and that if $k$ is some arbitrary constant, then $kp(x)\in S$ Is this all that is needed to show that the set of polynomials of degree less than or equal to 3 is a subspace of $P^3$?
-Is the dimension 4 since it is polynomials $<=3$?
Best Answer
The dimension is $3$. There are a number of ways of seeing this.
The dimension of $S$ is less than $4$, for $S$ is not all of $P_3$. The polynomials $x-7$, $(x-7)^2$, and $(x-7)^3$ are in $S$, and it is not hard to show they are linearly independent. So the dimension of $S$ is $\ge 3$. Since it is $\ge 3$ and $\lt 4$, it is exactly $3$.
More abstractly, the mapping $T$ that takes $p(x)$ to $p(7)$ is linear with kernel $S$. But the image of $T$ has dimension $1$, so the kernel has dimension $3$.