Determining if all polynomials of the form $a_{0}+a_{1}x$ is a subspace of $P_{3}$

Question

As the titles states I have to determine weather or not polynomials of the form $a_{0}+a_{1}x$ is a subspace of $P_{3}$, the polynomials of the form $a_{0}+a_{1}x$ have $a_{0}$ and $a_{1}$ as real numbers.

So since the polynomial is of the third degree, the entire polynomial would look like:

$a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3$

and the crux(I assume) is that I can pick any numbers for $a_{0}$ and $a_{1}$ to prove the that this is indeed a subspace of $P_{3}$.

Since subspaces with polynomials are equivocally described in my book, I would like to know when testing the axiom "closure under addition" for verifying if this polynimal is a subspace of $P_{3}$

adding two polynomials:

$(a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3) + (b_{0}+b_{1}x+b_{2}x^2+b_{3}x^3$)

I could substitute:

$a_{0} =1 $ $b_{0} =1$ and $a_{1} = 2$ $ b_{1} = -2$

Meaning that I would get:

\begin{align}(1+2x+a_{2}x^2+a_{3}x^3) + (1-2x+b_{2}x^2+b_{3}x^3) =(1+1) +(b_{2}+a_{2})x^2+(b_{2}+a_{2})x^3 \end{align}

Would this still be considered to be the subspace of $P_{3}$ even though the polynomial have't retained it's full length, however it has kept the same degree?

Help would be greatly appriciated!

Answer 1

To test closure under addition, you want to already start with polynomials in the alleged subspace. For example, $$(a_0+a_1x)+(b_0+b_1x)=(a_0+b_0)+(a_1+b_1)x$$ This is still in the same form (meaning it has no higher degree terms), so the subset is indeed closed under addition.

Answer 2

Step $1$: check that $Q=\{a_0+a_1x | a_0, a_1 \in \mathbb{R}\}$ is a subset of $P_3$.

This is true since we can write $a_0+a_1x=a_0+a_1x+0x^2+0x^3$.

Step $2$: check closure under addition:

$$(a_0 + a_1 x )+(b_0+b_1)x=(a_0+a_1)+(a_1+b_1)x$$

Note that $a_0+a_1, b_0+b_1 \in \mathbb{R}$.

• I will leave checking of closure under scalar multiplication as an exercise.

• Remark: you have to work with general entries rather than specific examples.