Question

which is subspace of p 2 with the usual operations?
a) span (1, x2)
b) (1+ax :a E R)
c) (a + b x2 : a, b E R)
d) (a: a E R)

Answer 1

Are there multiple answers possible?

To show that you have a subspace, you apply the subspace test... that is you check that the subspace (denoted by \(S\)) is a non-empty subset of your vector space (\(P_2\)). Furthermore, if 2 polynomials \(p_1(x)\) and \(p_2(x)\) satisfy

\(p_1(x)+p_2(x) \in S\) [i.e. it is closed under addition]
AND
if it satisfies \(cp(x) \in S, \text{for } c\in \mathbb{R}\) [i.e. closed under scalar multiplication],

then \(S\) will be a subspace of \(P_2\)

To verify if the subspace is non-empty, we usually check if the "zero element" is in \(S\), which is just the polynomial \(p(x)=0\). This guarantees the property \(cp(x) \in S\) to be satisfied.
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Let's try b):

Is \(p(x)=0\) in \(S\)? No, because \(p(x)=1+ax\) can never be 0 (even if you set a=0, you will get \(p(x)=1\). This already tells us that it can't be a subspace.

Let's try c):
Is \(p(x)=0\) in \(S\)? Yes, since \(p(x)=a+bx^2\) and you set \(a=0, b=0\), this gives \(p(x)=0\). We can set a=0 and b=0 since a and b are just real constants and can be anything in \(\mathbb{R}\).


To verify if closed under addition:
Say \(p_1(x)=a_1+b_1x^2\) and \(p_2(x) = a_2+b_2x^2\) where \(a_1, b_1, a_2, b_2 \in \mathbb{R}\)
Then, \(p_1(x)+p_2(x)=(a_1+b_1x^2) + (a_2+b_2x^2) = (a_1+b_1) + (a_2+b_2)x^2 \in S\)

because \((a_1+b_1) \in \mathbb{R}\) and \((a_2+b_2) \in \mathbb{R}\). If it doesn't seem clear, you can always replace \((a_1+b_1)\) with some general constant, say \(\alpha\) and \((a_2+b_2)\) with \(\beta\). So \(p_1(x)+p_2(x)=\alpha+\beta x^2 \in S\), because you get exactly the same "polynomial form" that is in \(S\).

To verify if closed under scalar multiplication:
Say \(p(x)=a+bx^2\) where \(a, b \in \mathbb{R}\) and introduce a constant \(c \in \mathbb{R}\).
Then, \(cp(x)=c(a+bx^2)=ca+(cb)x^2\) where \(ca \in \mathbb{R}\) and \(cb \in \mathbb{R}\). Hence, \(cp(x) \in S\). [Again, you can just replace ca by some constant, and replace cb by some constant)].