Finding a basis of p2

I have a vector space $$W=ax^2+bx+c$$ where $$a-b=0$$.
How would I find a basis for this? And also I need to state the dimension of $$W$$.

I also want to find a basis for Polynomial of degree 2 that contains my answer I have for The basis of $$W$$ as a subset?

• any polynomial of that specification will be a linear combination of $1$ and $x^2+x$, and dimension is the number of basis vectors Jan 16 '20 at 20:15
• How would I work that out then to find what you got for the basis. (My question is different to this one but wanted an example to work off) Jan 16 '20 at 20:23
• elements of $W$ are $ax^2+bx+c$ with $a=b;$ i.e., $ax^2+ax+c;$ i.e., $a(x^2+x)+c(1)$ Jan 16 '20 at 20:32
• So if I had a-2b=0 I would rearrange it to get it in terms of a? Jan 16 '20 at 20:42

Let $$\mathcal P_2$$ be the vector space of polynomials with real coefficients of degree at most $$2$$ and $$W\subset \mathcal P_2$$ the subspace $$\{ax^2+bx+c:a-b=0\}$$. Then an element of $$W$$ is of the form $$ax^2 + ax + c$$, for $$a,c\in\mathbb R$$. Since we can write $$ax^2 + ax + c = a(x^2+x) + c$$, it is clear that $$(x^2+x, 1)$$ is a basis for $$W$$, and hence $$\dim W=2$$.
• So if I has a-2b=0 would it be the same but ($x^2$+ $x/2$, 1) Jan 16 '20 at 21:31
• $a-2b=0$ if and only if $b=\frac12 a$, so yes. Jan 16 '20 at 21:56