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?

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    $\begingroup$ any polynomial of that specification will be a linear combination of $1$ and $x^2+x$, and dimension is the number of basis vectors $\endgroup$ Jan 16 '20 at 20:15
  • $\begingroup$ 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) $\endgroup$
    – Ellie
    Jan 16 '20 at 20:23
  • $\begingroup$ elements of $W$ are $ax^2+bx+c$ with $a=b;$ i.e., $ax^2+ax+c;$ i.e., $a(x^2+x)+c(1)$ $\endgroup$ Jan 16 '20 at 20:32
  • $\begingroup$ So if I had a-2b=0 I would rearrange it to get it in terms of a? $\endgroup$
    – Ellie
    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 $\dim W=2$.

  • $\begingroup$ So if I has a-2b=0 would it be the same but ($x^2$+ $x/2$, 1) $\endgroup$
    – Ellie
    Jan 16 '20 at 21:31
  • $\begingroup$ $a-2b=0$ if and only if $b=\frac12 a$, so yes. $\endgroup$
    – Math1000
    Jan 16 '20 at 21:56
  • $\begingroup$ Please could you take a look at the bit I added too! $\endgroup$
    – Ellie
    Jan 17 '20 at 8:58

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