Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $f_C$.

share|cite|improve this question
    
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
up vote 1 down vote accepted

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

share|cite|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.