My Math Forum f(φ(x))=φ(2)χ

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 January 11th, 2018, 11:05 AM #1 Member   Joined: Apr 2014 From: Greece Posts: 58 Thanks: 0 f(φ(x))=φ(2)x I saw an excercise with this weird looking linear transformation and I would like some help to find its matrix. $\displaystyle f(g(x))=g(2)x$ Here's my thought: So let $\displaystyle B=\left \{ u_{1},u_{2},u_{3}\right \}=\left \{ (x^2+x+1),(x),(1)\right \}$ be a basis of $\displaystyle \mathbb{R}_{2}[x]$ then: $\displaystyle f(u_{1})=(2^2+2+1)x=7x$ $\displaystyle f(u_{2})=2x$ $\displaystyle f(u_{3})=1$ So the matrix of $\displaystyle f$ in respect of $\displaystyle B$ is \begin{pmatrix} 0 & 0 & 0\\ 7 & 2 & 0\\ 0 & 0 & 1 \end{pmatrix} Is this correct? Or anywhere near correct? :P Never seen anything similar... Last edited by Vaki; January 11th, 2018 at 11:39 AM.
 January 12th, 2018, 02:46 PM #2 Member     Joined: Aug 2011 From: Nouakchott, Mauritania Posts: 85 Thanks: 14 Math Focus: Algebra, Cryptography Good evening ! Why $\displaystyle f\left(u_3)\right)(x)=1$ ? I think you forgot multiplying by $\displaystyle x$ and it should be $\displaystyle f\left(u_3)\right)(x)=1x=x$. And why you used that basis rather than using the canonical basis $\displaystyle \{1,x,x^2\}$ ? Anyway, it won't change a lot by we usually use the canonical basis. Thanks from topsquark

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