You are Here: Home >< Maths

# I have a question about vectors spaces... watch

1. 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.

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?

2. 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 " for an appropriate value of n" bit comes into this; surely they're all in ?
3. (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 " for an appropriate value of n" bit comes into this; surely they're all in ?
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)
4. (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.
5. (Original post by Slumpy)
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.
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.
6. (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.
I've given a counteracting +ve (and since "mine's bigger than yours", he should now be quids-in on the transaction!).

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: April 11, 2011
The home of Results and Clearing

### 1,060

people online now

### 1,567,000

students helped last year
Today on TSR

### University open days

1. Keele University
Sun, 19 Aug '18
2. University of Melbourne
Sun, 19 Aug '18
3. Sheffield Hallam University
Tue, 21 Aug '18
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams