Question

Theorem2 : If M is an m-dimensional subspace of an n-dimensional vector space V, then V/M has dimension n-m
The problem is: Consider the quotient spaces obtained by reducing the space P of polynomials modulo various subspaces. If M = P_n is P/M finite-dimensional? What if M is the subspace consisting of all even polynomials? What if M is the subspace consisting of all polynomials divisible by x_n (where x_n(t) = t^n)?
My question is how to make conclusion when P is infinite-dimensional which is not indicated in the theorem? Please, help

Answer 1

@e.mccormick

Answer 2

It's kind of English!!! :(

Answer 3

Is P the space of all Polynomials?

Answer 4

yes

Answer 5

Is P\M is the space of all Polynomials without the members in M?

Answer 6

yes

Answer 7

What exactly are the members of M?

Answer 8

M = P_n, we symbolize the dim (M) = m in general case
for example, if n = 4 , then M = {P_4}

Answer 9

So all we know is that the dimension of M is 4? It must have polynomials if we are removing them from the space P.

Then is M=P_n the space of all polynomials of dimension n?

Answer 10

yes,

Answer 11

Ok. So we have polynomials in P which include polynomials with infinite dimension. We remove from this set all polynomials of a certain finite dimension.

The question then is P\M finite dimensional?

Answer 12

yes

Answer 13

If I have all the natural numbers in my set. And each natural number represents the set of all polynomials of the number. For example, n=3 in this set is an index representing the set of all polynomials of dimension 3.

If from the natural numbers I remove a finite number of them, say 4 of them, is the number of members in my set still infinite or has it become finite?

It seems like this represents your problem.

If this is the case, if I remove a finite number of elements from an countably infinite set then you still have an infinite number of elements.

So P\M can not be finite dimensionally.

Answer 14

So, I should write out the whole thing as a logic for my conclusion?

Answer 15

If you wanted to prove

Let \(P_n\) represent all polynomials of dimension n
Without loss of generality, you can remove all polynomials from P of dimension 1 (i.e. remove)
You now have the set \(A=P\backslash P_1\)
The members of this new (A) set now contain \(P_2,P_3\cdots\)
Now, we want to form a 1-1 correspondence with the Natural numbers so that we can show that this new set is still of infinite dimension
So we let the elements of our set be the union of the elements \(A_1=P_2,A_2=P_3,\cdots\)
Where we assign the index of 1 to the 1st member of A corresponding to the Polynomials P_2 and the index of 2 to P_3 and so on
This correspondence shows that our new set A is of the same dimension as P

Answer 16

Second line of my proof-sketch should say
*(i.e. remove P_1)

Answer 17

Got what you mean. Thanks a lot. :)

Answer 18

You have the same situation if you remove an infinite number of polynomials from P as in removing the even polynomials.

You just assign A_1 to P1, A_2 to P_3 and so on, showing a 1-1 correspondence with the Natural numbers.

Answer 19

Ok, do homework now!! tomorrow is due day :)