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I have been reading through Linear Algebra Done Right by Sheldon Axler. The book defines an operator as a linear map from a vector space to itself. It then considers at another part of the book the operator of differentiation and states that the derivative of any polynomial of degree at most $4$ is a polynomial of degree at most $4$. I assume this means that this set polynomials is invariant under the differential operator. My crux with this is that wouldn't the differential operator map the vector space of polynomials of at most degree $4$ to the set of polynomials of at most degree $3$?

Apologies if this question does not make sense.

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    $\begingroup$ You are correct, but to put it differently the image of the differentiation is polynomials of degree at most 3, which is a perfectly decent subspace of polynomials of degree at most 4, the codomain. $\endgroup$ – Jason Knapp Sep 23 '14 at 12:57
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Nothing you said was wrong. Differentiation is a linear operator that transforms (one-variable) polynomial of degree at most $n$ into polynomial of degree at most $n-1$. Since a polynomial of degree at most $n-1$ is also a polynomial of degree at most $n$, you can say that this set is invariant under the differential operator.

If the dimension is your problem, remember that not every linear map is onto, in fact the differentiation is a singular operator on the space of polinomial, since $\frac{d}{dx}1=0$

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  • $\begingroup$ In fact, on the space of polynomials of degree at most $n$, differentiation is a nilpotent operator, since $\left(\frac{d}{dx}\right)^{n + 1} = 0$. This is not true on the space of all polynomials (at least, not when the underlying field has characteristic zero). $\endgroup$ – Travis Willse Sep 23 '14 at 13:17

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