Let D be the differentiation operator on P 3 , and let S = { p ∈ P 3 | p ( 0 ) = 0 } Show that (a) D maps P 3 onto the subspace P 2 , but D : P 3 → P 2 is not one−to−one. (b) D : S → P 3 is not one−to−one but not onto.
Let D be the differentiation operator on P 3 , and let S = { p ∈ P 3 | p ( 0 ) = 0 } Show that (a) D maps P 3 onto the subspace P 2 , but D : P 3 → P 2 is not one−to−one. (b) D : S → P 3 is not one−to−one but not onto.
Solution Summary: The author explains that D maps P 3 onto the subspace P 2 but is not one-to-one.
Let D be the differentiation operator on
P
3
, and let
S
=
{
p
∈
P
3
|
p
(
0
)
=
0
}
Show that (a) D maps
P
3
onto the subspace
P
2
, but
D
:
P
3
→
P
2
is not one−to−one. (b)
D
:
S
→
P
3
is not one−to−one but not onto.
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