Consider $\mathbb{P}_n[x]$ and $\mathbb{P}[x]$ over $\mathbb{R} $.

Are the following statements true / false?

a) $\{ p(x)\mid p(x) = a x^2 , a \in \mathbb{R} \}$ is a subspace of $\mathbb{P}_3[x]$

b) $\{p(x)\mid p(x) = a + x^2 , a \in \mathbb{R} \}$ is a subspace of $\mathbb{P}_3[x]$

My attempt: both a) and b ) are false.

For a) I take $p(2x) =a4x^2 \neq 2p(x)$

For b) $p(2x) = a + 4x^2 \neq 2p(x)$

Is it correct?

Any hints or solutions?


Recall what a subspace must satisfy : it must be closed under addition and scalar multiplication, and contain the zero element (and must be a subset of the vector space you are checking it is a subspace of, but this is clear here).

In none of the conditions does $p(2x)$ come into play. You should explain from where the idea that $p(2x) = 2p(x)$ must be true, came to you. The check performed is incorrect.

Note that $p(2x)$ is not the scalar multiplication of $2$ with $p(x)$ : it is the polynomial $p$, applied to the input $2x$. This is something that could differ wildly from $p$ : for example, if $p(x) = 3x^2 + 5x + 6$ then $p(2x) = 12x^2 + 10x + 6$, which is very different from $p(x)$, and will not be $2p(x)$ in general.

For the first example, remember the three points : closed under addition, closed under scalar multiplication, and zero is a member.

Is it closed under addition? $ax^2+bx^2 = (a+b)x^2$.

Is it closed under scalar multiplication? $c(ax^2) = (ca)x^2$.

Does it contain zero? $0 = 0(ax^2)$.

Hence, it is a subspace.

The second set does not contain zero, since $a+x^2$ has degree two always, while $0$ has no degree. So, $0 = a+x^2$ cannot happen, hence the second is not a vector space.

  • $\begingroup$ thanks actoh,,. $\endgroup$
    – jasmine
    Jun 29 '18 at 13:42

A is true. Let $p_1(x) = a_1 x^2$ and $p_2(x)= a_2 x^2$ are two elements of the given set, where $a_1,a_2 \in \mathbb R$ then $p_1(x)+p_2(x)= (a_1+a_2)x^2$, which is also belongs to the same set( because $a_1+a_2 \in \mathbb R)$. So it is closed under addition. And let $k \in \mathbb R$ then $kp (x)=(k.a) x^2$, so it is closed under scalar multiplication. So, it is subspace.

B is false. Multiply $x^2$ by 5, which is $5x^2$ and it doesn't belong to the same set. So, it is not subspace.


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