# linear-algebra bases polynomials

the answer that was provided was great help thank you, but i had a similar type of question to this once and i used the same method but was marked incorrectly.. it didnt state to compute the inverse and was marked wrong. is there another way of doing it? thanks in advance

Let $B := [p_0, p_1, p_2]$ denote the natural ordered basis for $P_2(\mathbb R)$, the vector space of real polynomial functions of degree less than or equal to $2$. Define $f_1, f_2, f_3\in P_2(\mathbb R)$ by $f_1(x) = 1 − x$, $f_2 = x − x^2$ and $f_3(x) = 1 + 2x + x^2$. Define $C := [f_1, f_2, f_3]$. Verify that $C$ is an ordered basis for $P_2(\mathbb R)$. Compute the change of coordinates matrix $A$ which converts $B$-coordinates to $C$-coordinates. Define $f \in P_2(\mathbb R)$ by $f(x) = 3 − 4x + 2x^2$. Compute $f_C$.

-
Unfortunately, there are two "natural ordered basis" choices for $P_2(\mathbb{R}$: $[1,x,x^2]$, and $[x^2,x,1]$. You should specify which you mean. – Arturo Magidin Jun 29 '12 at 14:17

Assuming you mean $\,B:=\{1,x,x^2\}\,$ , and since $$a(1-x)+b(x-x^2)+c(1+2x+x^2)=0\,\,,\,a,b,c\in\mathbb R\Longrightarrow$$ $$\Longrightarrow(a+c)+(b-a+2c)x+(c-b)x^2=0\Longrightarrow b=c=-a\,\, (\text{first and last coefficients})\,\,,$$ $$\,b-a+2c=-a-a-2a=-4a=0\Longrightarrow a=b=c=0$$ and thus $\,C\,$ is a basis.
Since \begin{align}1-x&=&1\cdot 1&+&(-1)\cdot 1&+&0\cdot x^2\\x-x^2&=&0\cdot 1&+&1\cdot x&+&(-1)\cdot x^2\\1+2x+x^2&=&1\cdot 1&+&2\cdot x&+&1\cdot x^2\end{align}
the wanted matrix is $$A=\begin{pmatrix}1&0&1\\\!\!\!\!-1&1&2\\0&\!\!\!\!-1&1\end{pmatrix}^{-1}=\frac{1}{4}\begin{pmatrix}3&-1&-1\\1&\;\;1&-3\\1&\;\;1&\;\;1\end{pmatrix}$$
Finally, since $$f(x)=3-4x+2x^2\stackrel{coord. wrt B}\longrightarrow \begin{pmatrix}3\\\!\!\!\!-4\\2\end{pmatrix} \,\,,\,\text{we get}$$ $$[Af]=\frac{1}{4}\begin{pmatrix}3&-1&-1\\1&\;\;1&-3\\1&\;\;1&\;\;1\end{pmatrix}\begin{pmatrix}3\\\!\!\!\!-4\\2\end{pmatrix}=\frac{1}{4}\begin{pmatrix}11\\\!\!\!\!-7\\1\end{pmatrix}$$ so $$\,f_C=\frac{11}{4}(1-x)-\frac{7}{4}(x-x^2)+\frac{1}{4}(1+2x+x^2)$$