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Let $T \in \operatorname{End}(\mathbb{R}[X])$

How to think of inverse of $T$ ?

I thought about it in two ways:

$a)\quad \exists U\in \mathbb{R}[X]: Tf\cdot Uf = 1$,

$b)\quad U\left(Tf\right) = f.$

Is it true that neither $T$ nor $U$ can be matrices, because they are in $\mathbb{R}[X]$ ?

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It is a linear operator on $\Bbb R[X]$, not in that space. And for linear operators, the multiplication structure of polynomials is ignored, so it cannot be a). Option b) is better, but is incomplete. First of all you need to say this condition $U(T(f))=f$ should hold for any $f\in\Bbb R[X]$ (in spite of appearances, $f$ is not a function but a polynomial). But even then this is only half of the requirement. One should also have $T(U(p))=p$ for every $p\in\Bbb R[X]$. Unlike in finite dimension, that second condition is not a consequence of the first. For instance, if $T$ is the operation of multiplying by some fixed polynomial, say by $2X^3-X+7$, and $U$ is the operation of taking the Euclidean quotient by that polynomial (ignoring any remainder in the division), then both are linear operations and $U(T(p))=p$ holds for any polynomial $p$, but $T(U(p))$ does not: it only holds when the Euclidean division in computing $U(p)$ produces no remainder.

And you cannot have (finite) matrices because the space has infinite dimension.

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  • $\begingroup$ Why is multiplication structure ignored for linear operators ? And how to view $T$ on $\mathbb{R}[X]$. For, if $T \in Mat_{n,k}(K)$ then $T$ would be some matrix. So in this case, is $T$ a polynomial ? $\endgroup$ – flowian May 15 at 7:32
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    $\begingroup$ A famous example of a linear operation $T \in End(\mathbb{R}[X])$ is the derivative operator, i.e $Tf = f'$. You should think of polynomials as vectors. The operator takes a vector as input and returns a different vector as output, just as a matrix would do. You could think about how you would write the derivative operator as a matrix. The first thing you need is a basis for your vectorspace $\mathbb{R}[X]$. The second thing you run into is that your matrix has dimension infinity by infinity (so some people might refuse to call it a matrix). You can still compute its entries though. $\endgroup$ – Vincent May 15 at 7:46
  • $\begingroup$ (Ok I realize now that in the context of the current question the differentiation operator is a bad example as it doesn't have an inverse, but I just wanted to show what operators on this space could look like) $\endgroup$ – Vincent May 15 at 7:48
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    $\begingroup$ @flowian The multiplication structure is ignored because multiplication of vectors is not something provided in a vector space (thought multiplication by scalars is provided). So if a multiplication is nonetheless defined, as it is for polynomials, then viewing $\Bbb R[X]$ as a vector space ignores that structure. And linear operators are defined in terms of the vector space structure (only). $\endgroup$ – Marc van Leeuwen May 15 at 7:51
  • $\begingroup$ @Vincent It is very good example actually. The question arose from problem where T is differentiation operator ($Tf = f + f'$) $\endgroup$ – flowian May 15 at 7:52

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