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I'm quite new to linear algebra and need help finding the way to solve a question that I know the answer to.. How to think? The linear operator $T(f(x))=xf'(x)+f(2)x+f(3)$ and I want to find the ordered basis such that $[T]B$ is a diagonal Matrix. I know that $T(1)=x+1$, $T(x)=3x+3$ and $T(x^2)=2x^2+4x+9$, but how to get there? I dont understand what $f(2)x$ means and what is this f.example when we want to solve $T(x)$?

Thanks for your help! :)

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    $f(2)x$ means the function that is $cx$ with $c$ chosen as $f(2)$. Your linear operator operates at a higher layer of functions. That is it takes a function as input and produces another function as output. Imagine differentiation, it takes the finction $\sin x$ as input and outputs another function $\cos x$. – 2017-02-27
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    Ok, thanks! But how do you then solve: T(1)=? T(x)=? T(x^2)= ? – 2017-02-27
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    If you add a dollar sign (`$`) either side of your maths equations, it wil format it nicely - i.e. `$T(x^2)$` becomes $T(x^2)$ rather than just T(x^2) – 2017-02-27
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    Can you tell us what the space is that $T$ is acting on? – 2017-02-27

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There is some information missing, especially what the domain of your operator is, in other words: What functions are you allowed to use for $f(x)$?

This could in theory be an function $f: \mathbb R \rightarrow \mathbb R$, like $f(x)=\sin(x)$, but from your "known answer" it seems to be that $f(x)$ is restricted to polynomials, so I'll assume that.

$T(1)$ is simply the function that $T$ produces when you put in the constant function $f(x)=1$ for all $x \in \mathbb R$. We have in that case $f'(x)=0$ for all $x \in \mathbb R$ and $f(2)=f(3)=1$ and the formula

$$T(f(x))=xf'(x)+f(2)x+f(3)$$ results in

$$ T(1) =x\cdot 0 + 1\cdot x + 1 = x+1.$$

If you use $f(x)=x$ for all $x \in \mathbb R$, you get $f'(x)=1$ for all $x \in \mathbb R$ and $f(2)=2, f(3)=3$ and hence

$$T(x)=x\cdot 1 + 2\cdot x + 3 = 3x+3.$$

You should be able to verify $T(x^2)$ on your own now, and maybe find the general formula for $T(x^n)$.

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    This does not answer the question. The OP is looking for a basis to diagonalize its operator. – 2017-02-27
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    The OP seems to be a beginner, not knowing what "$f(2)x$" means. He/she seems unaware where the formulas he/she "knows" as the answers come from. I tried to explain how they were derived. Maybe I should have added that the $T(x^n)$ are the basis for the image vector space that OP is supposed to find. But, no disrespect to the OP, I doubt that he/she understands what that means, judging from the 'beginner level' questions he/she asked. – 2017-02-27