Linear Algebra Subspaces and dimension 

Let 
    $$W_1 = \lbrace\; ax^4 + bx^3 + ax^2 + cx \in P_4(\mathbb{R}) | \ a, b, c ∈ \mathbb{R}\;\rbrace$$ 
    and 
    $$W_2 =  \lbrace\; 0x^4 + ax^3 − ax^2 + 2bx + a \in  P_4(\mathbb{R}) | \ a, b ∈ \mathbb{R} \;\rbrace.$$

(a) Show that $W_1$ and $W_2$ are subspaces of $P_4(\mathbb{R})$. 
(b) Find bases for $W_1$ and $W_2$. 
(c) Determine the dimensions of $W_1$ and $W_2$.

So for the subspace part, can we just show 
$$r(a,b,a,c) = (ra, rb, ra, rc)\in P_4(\mathbb{R})$$
and
$$(a, b, a , c ) + (a, -a, 2b, a) = (2a, b-a, a+2b, c+a) \in P_4(\mathbb{R})?$$
For the bases, my answer is:
$$W_1=\langle\; (1,0,1,0), (0,1,0,0), (0,0,0,1)\;\rangle$$
and 
$$W_2 = \langle\; (1,-1,0,1), (0,0,2,0) \;\rangle $$ 
so the dimensions are 3 and 2 respectively.
I just wanted to make sure that my work is correct. Thanks in advance
 A: I am going to assume the space $P_4(\mathbb{R})$ is the set of polynomials of degree $4$ or less, with the regular polynomial addition. Also, I assume you are working with $\mathbb{R}$ as your scalar field.


*

*To begin with, saying that $W_1$ and $W_2$ are subspaces of $P_4(\mathbb{R})$ just means that $W_1\subset P_4(\mathbb{R})$ $W_2\subset P_4(\mathbb{R})$. You probably mean linear subspaces. If that is the case, you have three conditions to work with, as you may alread know. First, $0\in W_1$ by taking $a=b=c=0$. Secondly, if you have $p_1,p_2\in W_1$, then they are of the form
$$p_1=a_1x^4+b_1x^3+a_1x^2+c_1x$$
$$p_2=a_2x^4+b_2x^3+a_2x^2+c_2x$$
with the coefficient taken in $\mathbb{R}$. For any $\lambda\in\mathbb{R}$, you have
$$p_3=p_1+\lambda p_2=(a_1+\lambda a_2)x^4+(b_1+\lambda b_2)x^3+(a_1+\lambda a_2)x^2+(c_1+\lambda c_2)x^4.$$
Since you can write $p_3$ as the required form, $p_3\in W_1$, as wanted. I will leave you $W_2$ to do.

*For the basis, your answer is wrong. Be careful on your representation of polynomials as a tuple. One can show that 
$$ \lbrace\; x^4+x^2,x^3,x\;\rbrace $$
is a basis for $W_1$. With your notation, this gives
$$\lbrace\; (1,0,1,0,0),(0,1,0,0,0),(0,0,0,1,0)\;\rbrace $$
as a basis. Don't forget you need to show that every element of $W_1$ is a linear combination of the basis by explicit calculations. Again, I leave you $W_2$.

*The answer you gave for the dimension is correct. It is the size of the basis.


I hope this helps you.
