Question

Use the differentiation operator on \(P_n\) to show that the range and the null space of a linear transformation need Not be disjoint
Please help.

Answer 1

is \(P_n\) the set of polynomials of degree <= n ?

Answer 2

yes, please

Answer 3

first of all
\[\large P_n=\langle1,x,x^2,\dots,x^n\rangle. \]
if \(D:P_n\to P_n\) is given by \(D(p(x))=d p(x)/dx\), then
\[\large \ker D=\langle 1\rangle \]
and
\[\large \text{im}\ D=\langle1,2x,3x^2,\dots,nx^{n-1}\rangle=
\langle1,x,x^2,\dots,x^{n-1}\rangle \]
therefore
\[\large \ker D\cap\text{im }D=\langle1\rangle\neq\emptyset \]

Answer 4

what does im (D) mean? I don't understand why im (D) = <1,2x,3x^2,....,nx^n-1> = .....
how and why? if it stop there, so I understand that is the set of target after take derivative, but you let it = < basis of P_n>
it confused me.

Answer 5

i don't know how u call it in english. "im" stands for image (imagen in spanish).

Answer 6

oh, you mean range (D)

Answer 7

yes

Answer 8

the second line P_n=<...> is the cannonical base of Pn.

Answer 9

how they are equal?

Answer 10

i see. i guess u use the notation "span" so
\[\large \langle1,x,\dots,x^n\rangle=\text{span}\{1,x,\dots,x^n\} \]

Answer 11

<1,x,......x^n> is basis of P_n (not span) but how it is = range (D)

Answer 12

oh, I got what you mean now. Thank a ton

Answer 13

sure. u r welcome