Let $D$ be the Differentation operator of the of polynomials over $R.$ Prove that there is no polynomial $g(t),$ such that $g(D)=T_0.$

But characteristic polynomials satisfies it's operator.

I dnt understand the phenomenon.


The space of polynomials is infinite-dimensional. So there is no characteristic polynomial, and no Cayley-Hamilton theorem.

If $g$ is a polynomial of degree $n$, then $g(D)$ is the differentiation of order $n$. If you take a polynomial $p$ of degree $n+1$, then $g(D)p\ne0$, hence $g(D)\ne0$.

On the other hand, if you consider the space of polynomials of maximum degree $n$, which is finite-dimensional, then $g(t)=t^{n+1}$ is the minimal polynomial of $D$.


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