Modular arithmetic on vector spaces

  • #1
gruba
206
1

Homework Statement


Let [itex]U[/itex] is the set of all polynomials [itex]u[/itex] on field [itex]\mathbb F[/itex] such that [itex]u(3)=u(-2)=0[/itex]. Check if [itex]U[/itex] is the subspace of the set of all polynomials [itex]P(x)[/itex] on [itex]\mathbb F[/itex] and if it is, determine the set [itex]W[/itex] such that [itex]P(x)=U\oplus W[/itex].

Homework Equations


-Polynomial vector spaces
-Subspaces
-Modular arithmetic

The Attempt at a Solution


[itex]U=\{u(x):u(x)\mod (x-3)=0 \land u(x)\mod (x+2)= 0\}[/itex]

[itex]U[/itex] is the subspace of [itex]P(x)[/itex] iff

[itex]1)[/itex] [itex]\forall u_1,u_2\in U\Rightarrow u_1+u_2\in U[/itex]

[itex]2)[/itex] [itex]\forall u\in U,\forall \alpha\in \mathbb F\Rightarrow \alpha u\in U[/itex]

How to check if [itex]U[/itex] is the subspace of [itex]P(x)[/itex]?

Assuming [itex]U[/itex] is the subspace of [itex]P(x)\Rightarrow[/itex]

[itex]P(x)=U\oplus W=U+W \mod n[/itex]

where [itex]n[/itex] should be the total number of polynomials in [itex]U[/itex] and [itex]W[/itex].

This means that [itex]W[/itex] is the set of all polynomials [itex]u(x)[/itex] defined as
[itex]W=\{u(x): u(3)\neq u(-2)\neq 0 \lor u(3)\neq u(-2)=0\lor u(3)=u(-2)\neq 0\}[/itex]

Is this correct?
 
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  • #2
gruba said:

Homework Statement


Let [itex]U[/itex] is the set of all polynomials [itex]u[/itex] on field [itex]\mathbb F[/itex] such that [itex]u(3)=u(-2)=0[/itex]. Check if [itex]U[/itex] is the subspace of the set of all polynomials [itex]P(x)[/itex] on [itex]\mathbb F[/itex] and if it is, determine the set [itex]W[/itex] such that [itex]P(x)=U\oplus W[/itex].

Homework Equations


-Polynomial vector spaces
-Subspaces
-Modular arithmetic

The Attempt at a Solution


[itex]U=\{u(x):u(x)\mod (x-3)=0 \land u(x)\mod (x+2)= 0\}[/itex]
How does modular arithmetic come into this problem? In the problem statement, you say only that u(3) = 0 and u(-2) = 0.
gruba said:
[itex]U[/itex] is the subspace of [itex]P(x)[/itex] iff

[itex]1)[/itex] [itex]\forall u_1,u_2\in U\Rightarrow u_1+u_2\in U[/itex]

[itex]2)[/itex] [itex]\forall u\in U,\forall \alpha\in \mathbb F\Rightarrow \alpha u\in U[/itex]
Minor point, but ##\Rightarrow## is not appropriate in the above. It should be used when one statement implies another statement.##\forall u \in U## is not a statement (i.e., a sentence whose truth value can be determined).
gruba said:
How to check if [itex]U[/itex] is the subspace of [itex]P(x)[/itex]?

Assuming [itex]U[/itex] is the subspace of [itex]P(x)\Rightarrow[/itex]

[itex]P(x)=U\oplus W=U+W \mod n[/itex]

where [itex]n[/itex] should be the total number of polynomials in [itex]U[/itex] and [itex]W[/itex].

This means that [itex]W[/itex] is the set of all polynomials [itex]u(x)[/itex] defined as
[itex]W=\{u(x): u(3)\neq u(-2)\neq 0 \lor u(3)\neq u(-2)=0\lor u(3)=u(-2)\neq 0\}[/itex]

Is this correct?
 
  • #3
Any polynomial, u(x), of degree n, such that u(3)= u(-2)= 0 is of the form (x- 3)(x+ 2)v(x) where v is a polynomial of degree n- 2. To show this is a subspace of the space of all polynomials, you only need to show that this set is closed under addition and scalar multiplication- If u(x)= (x- 3)(x+ 2)v(x) and w(x)= (x- 3)(x+ 2)y(x), what can you say about u+ w? What can you say about au where a is a number?
 
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