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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 $f_C$.

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Please 1. Rewrite your question so it doesn't read like an order, 2. Tell us what you've done, what relevant facts you know, where you get stuck, etc., so we have a better idea how to help, 3. Learn how to use TeX so you get $x^2$ instead of x2, 4. If this is homework, add the "homework" tag. –  Gerry Myerson Jun 29 '12 at 13:03
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I edited your question to help you on your first steps for the "learn TeX"-part. You should work on the other two points yourself. –  Simon Markett Jun 29 '12 at 13:31
    
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
    
hi, really sorry if i made mistakes, im new to this. this was an exercise that my teacher gave to me. thanks you very much for answering my question. –  user34742 Jul 1 '12 at 15:45
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1 Answer 1

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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 $$\,f_C=\frac{11}{4}(1-x)-\frac{7}{4}(x-x^2)+\frac{1}{4}(1+2x+x^2)$$

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ive had a similar question to this before and i computed the inverse of A but it was marked wrong.. is there another way of doing it? –  user34742 Jul 28 '12 at 10:37
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