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January 11th, 2018, 12:05 PM   #1
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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 12:39 PM.
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January 12th, 2018, 03:46 PM   #2
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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.
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