What are considered vectors in P2?

For example, x^2, (1,1,1), 1, x+3, etc, which are vectors, and what are not?

Question

What are considered vectors in P2?

For example, x^2, (1,1,1), 1, x+3, etc, which are vectors, and what are not?

jamesj · Answer

By definition P2, is the set of polynominials of order 0, 1 and 2, considered as a vector space.  Hence the arbitrary member of P2 is

ax^2 + bx + c

where a, b, c are members of the field F over which we're making the polynominials.  If that bit confuses you because you haven't talked about fields yet, let's just say a, b and c are real numbers and constants.

anonymous · Answer

Hmm... but what confuses me is that someone also tells me that I can write them in tuples of 3, since there are 3 coefficients in P2. 

So, by definition, P2 also includes P1 and P0, where P0 is pretty much just all real numbers? And, yea, we haven't talked about fields yet, or I didn't pay enough attention to catch that Dx

jamesj · Answer

Well, there's a natural association of
$$P_2 \rightarrow \mathbb{R}$$
by
$$ax^2 + bx + c  \ \mapsto \ (a,b,c)$$

jamesj · Answer

$$P_2 \rightarrow \mathbb{R}^3$$

anonymous · Answer

... hm. So. 

A. 2x^2 +5
B. X^2
C. (1,0,0)
D. 1
E. 2x+3
F. |dw:1318888900619:dw|
G. x
H. None of the above.

A B C D E G are considered vectors of P2  Do I get the right idea?

anonymous · Answer

... and that drawing is supposed to be a vertical column of 1's, with parentheses, though I have no idea what that is either.

jamesj · Answer

ABDE are the only members here of P2.  The others are NOT members of P2.

anonymous · Answer

Oh.. what about G. How come that isn't a member of P2? If E is considered a member, why doesn't G work?  And, I'm still confused about the tuples of 3.  (x,x,x) are associated with P2, but not actually part of P2?

jamesj · Answer

sorry.  I didn't see G.  Yes, G as well.

I can associate the 2x2 matrix
a b
c d

with vectors in R^4, but that doesn't mean the R^4 vector (1,2,3,4) is a 2x2 matrix.

Same here.  Just because (1,2,3) is associated by my map with x^2 + 2x + 3, that doesn't mean (1,2,3) is a member of P2.

anonymous · Answer

Okay! Thank you. :3 That answers that part of the question.  And if I'm getting the right idea, the zero vector is 0, right?

jamesj · Answer

Definitely, because for any member of P2, ax^2 + bx + c, when we add the zero polynomial 0, we get back the original polynomial.

anonymous · Answer

Kay! Thank you!  This is a lot easier to understand than trying to wade through the textbook. Onto next problem.