Linear Algebra Subspaces and dimension

Question

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

Answer 1

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.

1. 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.
2. 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 $W_2$.
3. The answer you gave for the dimension is correct. It is the size of the basis.

I hope this helps you.