- #1
Niles
- 1,866
- 0
Homework Statement
I have a differentiation operator on P_3, and:
S = {p \in P_3 | p(0) = 0}.
I have to show that
1) D : P_3 -> P_2 is not one-to-one.
2) D: S -> P_3 is one-to-one.
3) D: S -> P_3 is not onto.
The Attempt at a Solution
For #1, I want to show that our differentiation operator is not one-to-one by looking at the matrix that represents P_3 relative to P_2:
[tex]A = \left( {\begin{array}{*{20}c}
2 \hfill & 0 \hfill & 0 \hfill \\
0 \hfill & 1 \hfill & 0 \hfill \\
\end{array}} \right)[/tex]
From this matrix, how do I show this?
2 + 3) In this case, S is [0; bx; ax^2]. Again, I will use the same approach as in #1, but what is the connection between the matrix A and the operator being one-to-one or onto?