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I have a question about vectors spaces...

Determine if the given set is a subspace of P_n for an appropriate value of n. Justify your answer.

All polynomials of the form p(t) = a + t^2, where a is in R.

My answer:
The given set is NOT a subspae of P_n because...

It is NOT closed under addition: If there are two polynomials, u(t) = b + t^2 and v(t) = c+t^2, then... u(t)+v(t) = b + t^2 + c + t^2 = (c+b) + (2)t^2 ? p(c+b)

Another way to show that p(t) is not a subspace is to show how the zero polynomial is not included: Because "a" can be any real number, it can equal to zero, but the coefficient next to t^2 is equal to 1 and therefore cannot equal zero.

Is my answer correct? Is there anything wrong with it?

Thank you in advance.

(edited 13 years ago)

Reply 1

nuodai

Your answer is fine. Saying that the zero polynomial isn't in it is enough, and so is saying that it's not closed under addition; you don't need to say both.

Although I can't see where the "

P_n

for an appropriate value of n" bit comes into this; surely they're all in

P_2

Reply 2

Artus

Original post by nuodai

Your answer is fine. Saying that the zero polynomial isn't in it is enough, and so is saying that it's not closed under addition; you don't need to say both.

Although I can't see where the "

P_n

for an appropriate value of n" bit comes into this; surely they're all in

P_2

Thank you...is it ok if you also check the way I show that it's not closed under addition?

"u(t)+v(t) = b + t^2 + c + t^2 = (c+b) + (2)t^2 which is not equal to p(c+b)

(edited 13 years ago)

Reply 3

Slumpy

Original post by Artus

Thank you...is it ok if you also check the way I show that it's not closed under addition?

"u(t)+v(t) = b + t^2 + c + t^2 = (c+b) + (2)t^2 which is not equal to p(c+b)

Not quite.
This much is right:
"u(t)+v(t) = b + t^2 + c + t^2 = (c+b) + (2)t^2"
Then all you need is that this isn't in the space, so is not of the form a+t^2(which it isn't). p(c+b) would be something entirely different, and unrelated.

Reply 4

Artus

Original post by Slumpy

Thank you for answering...I wanted to give you a thumbs up but accidentally clicked on the thumbs down...is there any way I can change it? I'm so sorry. :frown:

Reply 5

DFranklin

Original post by Artus

Thank you for answering...I wanted to give you a thumbs up but accidentally clicked on the thumbs down...is there any way I can change it? I'm so sorry. :frown:

I've given a counteracting +ve (and since "mine's bigger than yours", he should now be quids-in on the transaction!).