Why is the derivative map a linear transformation?

Question

My textbook defines a linear transformation as a linear map from a space into itself $t:V \rightarrow V$, so basically where the domain equals the codomain. It then goes on to say that the derivative map d/dx: $P_n \rightarrow P_n$ is a linear transformation:

$a_0 + a_1x + ... + a_nx^n \xrightarrow{d/dx} a_1 + 2a_2x + 3a_3x^2 + ... + na_nx^{n-1}$.

But how can this be? By losing the $a_0$, you've lost 1 component, so your dimension gets reduced by 1. How can this still be a linear transformation or am I missing something here?

What in the definition of linear transformation relies on the notion of dimension? — blargoner, Dec 06 '22 at 20:19
See answer below but anyway, your textbook is very guilty if it really tells you that a linear map is necessarily from a space to itself. — Anne Bauval, Dec 06 '22 at 20:25
Presumably, $P_n$ here is the space of polynomials of degree at most $n$, so a polynomial of degree $n-1$ is still in that same space. — Alex Jones, Dec 07 '22 at 06:58

score 9 · Answer 1 · answered Dec 06 '22 at 20:23

9

The derivative of a sum is the sum of the derivatives. If you multiply a function by a constant you multiply the derivative by the same constant. That makes differentiation linear on a vector space of differentiable functions, so in particular on a space of polynomials.

If your space is the space of polynomials of degree at most $n$ then the polynomial $x^n$ is not the derivative of anything, so it's not in the image of the differentiation operator. That's not a contradiction of any kind.

answered Dec 06 '22 at 20:23

Ethan Bolker

95,224
7
108
199

Okay, so my confusion was that I neglected to remember that a polynomial of degree x^n-1 is still in the space of polynomials with the degree x^n. Thanks! – Mezzoforte Dec 07 '22 at 13:54
You're welcome. You want degree at most $n$. – Ethan Bolker Dec 07 '22 at 13:58

Why is the derivative map a linear transformation?

1 Answers1