# Show $S$ is a subspace of $P_2$

Let $$P_2$$ be the set of real polynomials of degree $$\ge 2$$ in the variable $$x$$. That is $$P_2 = \{f(x) = a_0+a_1x+a_2x^2 | a_0, a_1, a_2 \in\mathbb{R}\}$$. We have shown that $$P_2$$ is a vector space under polynomial addition and scalar multiplication. Consider the subset $$S = \{f \in P_2 | f(1) = 0\}$$ (that is for example that $$x - 1$$ is an element of $$P_n$$). (1) Show the $$S$$ is a subspace of $$P_2$$ and (2) find a set that spans $$S$$.

I am just learning subspaces and spans so I am not quite sure about how to go about starting the solution to this problem.

• Is it closed under addition and scalar multiplication? And did you mean $P_2=\{f(x)=a_0+a_1x+a_2x^2...$? Sep 24, 2020 at 0:53
• Yes sorry, my formatting is bad. We know P is a vector space under polynomial additional and scalar multiplication and I did mean that formatting for P Sep 24, 2020 at 1:22
• You write, "degree at least 2," but your set is for degree at most 2. The polynomials of degree at least 2 do not form a vector space, since there is no zero element, and no closure under addition. Sep 24, 2020 at 1:50
• Are you listening, A person? Sep 25, 2020 at 12:25
• The problem was given to me by my teacher, I just copied how it was given to me. That is what caused me so much confusion but I chalked it up to him meaning at most not at least Sep 25, 2020 at 23:30

For (1) you need to show it is closed under summation and scalar multiplication, meaning that if you take $$f,g \in P_2$$ that satisfy it and an $$\alpha \in \mathbb{R}$$ then $$f+g$$ and $$\alpha f$$ satisfy it.
For (2) you need to find the solution to the linear system $$f(1)=0$$ for arbitrary $$f$$ and see how you can write the answer as a combination of two distinct vectors which you rewrite as polynomials, if you learnt about dimension/basis you need to find two distinct solutions to it and that will do it. Also $$P_2$$ is generally defined as polynomials with degree less than $$2$$ so you probably made a typo.