Question

Consider the vector space P and the subset V of P consisting of those vectors (polynomials) x for which:
1/ 2x(0) =x(1)
2/ x(t) = x(1-t)for all t
Why are they not vector spaces?
Please, help

Answer 1

@tkhunny

Answer 2

that's a tuff one :/

Answer 3

What are the Vector Space Properties? List them and prove them.

Answer 4

u, v in P , then u+ v in P
k \(\vec u\)= (ku1, ku2, ku3)
the def is quite simple but how to apply?

Answer 5

and we have a theorem said that P is a vector space, V is subset, we just prove those property, not the whole 10 axioms of Vector space

Answer 6

1) Closure

Can you find a vector in V that, when subjected to the given operations, results in something NOT in V?

Answer 7

Ok, let try
\(\vec u (t) = a+ a1x+ a2x^2 + a3 x^3+.......+.anx^n\)
\(\vec u(0) = a\)
\(vec u(1) = a+ a1+a2+....+an\)
then?? another one as v ?

Answer 8

Does it need to be one-to-one?
Anything about the Kernel?
Check it out!

Answer 9

@loser66 one way a subset could not be a vector space is if it did not contain the zero vector. The subspace must contain the zero vector to be a subspace. In other words, it must contain the origin. Do yours include the zero vector?

Answer 10

@ybarrap we have a theorem as I said above. if V is subset .....

Answer 11

The zero polynomial is included in P

Answer 12

Doesn't x(t) = x(1-t) for all t make x(t) a constant? Also, do 1/ and 2/ above define the subset V?

So condition 1 on V is 2x(0)=x(1) and condition 2 on V is x(t) = x(1-t)for all t?

x(0) could never be zero, because then x(1) would be zero and then so would all x, since x(t) is a constant.

Then so either x(t) is the zero vector or x(t) does not contain the zero vector, in which case it is not a subspace.

Answer 13

@ybarrap 1 and 2 are 2 separate problems.

Answer 14

@Loser66, it would be an advantage if you could write down an example of how the problem states that a vector would look like, reading the problem it makes it very confusing:

(polynomials) x for which:
1/ 2x(0) =x(1)"

I guess the 1 refers to the first problem, but what is the property \(2x(0)=x(1)\) supposed to mean? If it is a regular polynomial, then \(p(x)=a_0 + a_1x + a_2 x^2 + \dots + a_n x^n\) and the above definition would make no sense, it seems to me like the polynomial gets evaluated twice, like in a polynomial ring.

Besides all that, you're on the right track, with polynomials addition is rarely a problem, the zero vector however usually is. So check if the zero vector is within your subspace, also let me highlight that just because the zero vector is in \(P\) does by no means at all mean that it is in the subspace \(V \subset P\).

Answer 15

what is x (0)?

Answer 16

I attach the book page where the problem is
Problem 5b, and d @Spacelimbus

Answer 17

I have exactly the same problem from another book and the solution, but I don't understand, hehehe. let me post it. Please, explain me if you get something from the problem

Answer 18

How can we show that the zero vector exists when all we have are polynomials with the property 2x(0)=x(1)? I don't know the answer, but I think it is worth determining this basic requirement to be a subspace.

What do polynomials with this property look like?

Say x(t) = ax(t+1) + bx(t+3) + ..

So x(0) = ax(1) + bx(3) + ...

Then 2x(0) = 2ax(1) + 2bx(3) + ...

But we need only the polynomials where,

2x(0) = 2x(1)

But not the ones where, for example,

2x(0) = 2ax(1) + 2bx(3) + ....

In other words, 2x(0) = 2x(1) consists of all polynomials where

for t=...-2,-1,3,4,.... x(t) = 0

And we also need a=1. Maybe this is key.

Can we build a subspace with only these two non-zero elements?

I don't know if this approach is correct, but maybe this can help.

Answer 19

I think I get it
2x(0) = x(1)
\(X= (x(0), 2x(0), x(2),......,x(n)\\Y= (y(0), 2y(0) , y(2),....., y(m)) \)assume n<m
therefore \(X + Y = ((x(0)+y(0), 2 (x(0) +y(0), x(2)+y(2),...............x(n)+y(n), y(n+1),....ym)\) closed addition
and easy to check for multiplication
--> It's a vector space

Answer 20

I believe in the book it's pretty well explained @Loser66 , the above vector spaces are subspaces in \(\mathbb{C}\). It is mainly the notation that seems very confusing to me. I haven't seen such a notation for a polynomial before. If the way you have written down the polynomial is what they want with it, then you're right.

Answer 21

Just check that your multiplication holds true for scalars in \(\mathbb{C}\) as well.

Answer 22

Yes, I will. Thanks @Spacelimbus .

Answer 23

@loser66, your approach looks logical and correct. The polynomials you describe are difference equations that can be rewritten using the Laplace Transform in terms of powers that relate the shifts to powers in s. So working with these types of polynomials (I think) is equivalent to working with the power series directly.

Answer 24

*I meant z-transform not Laplace - http://en.wikipedia.org/wiki/Difference_equation#Solving_with_z-transforms