Dimension of a subspace of polynomials with certain coefficients

[Zorba]

Right so I've had an argument with a lecturer regarding the following:

Suppose you consider P_4 (polynomials of degree at most 4): A(t)=a_0+a_1t+a_2t^2+a_3t^3+a_4t^4

Now if we consider the subspace of these polynomials such that a_0=0,\ a_1=0,\ a_2=0}, I propose that the dimension of of this subspace is 2 (versus the dimension of P_4 which is 5. Am I incorrect in saying this?

Based on the answer to this I have a follow up question regarding a linear operator on P_n

 

[HallsofIvy]

Right so I've had an argument with a lecturer regarding the following:

Suppose you consider P_4 (polynomials of degree at most 4): A(t)=a_0+a_1t+a_2t^2+a_3t^3+a_4t^4

Now if we consider the subspace of these polynomials such that a_0=0,\ a_1=0,\ a_2=0}, I propose that the dimension of of this subspace is 2 (versus the dimension of P_4 which is 5. Am I incorrect in saying this?

Yes, any such polynomial can be written as 0(1)+ 0(x)+ 0(x^2)+ a_3 t^3+ a_4t^4 and so \{t^3, t^4\} is a basis.

Based on the answer to this I have a follow up question regarding a linear operator on P_n

Fire away!

 

[Zorba]

(Cheers)

Suppose we consider an operator B: A(t) \rightarrow tA(t) where A(t) \in P_4

Right, so my argument is that the result of this operation does not change the dimension, ie that dim(A(t))=dim(tA(t)) (basically I was trying to convince him that this was a linear operator). Am I missing something here?

He seemed to argue that since tA(t) now belongs to P_5 it now has an extra dimension, but I argue that since tA(t) has the form a_0t+a_1t^2+a_2t^3+a_3t^4+a_4t^5 and since a polynomial in P_5 would be A(t)=a_0+a_1t^1+a_2t^2+a_3t^3+a_4t^4+a_5t^5 that it does indeed have the same dimension...

 

[brian44]

You are correct, the range of B has the same dimension, however, B is not a linear operator, because a linear operator is a linear transformation from a space to itself, i.e. the same space, it must take P_4 -> P_4, but tA(t) takes P_4->P_5, although to a 2 dimensional subspace in P_5 it is no longer the same space. (E.g. now we have t^5 in the range of B, but by definition that does not exist in P_4, so the spaces are different).

 

[Zorba]

Ahha, I had (oddly in retrospect) thought that a linear operator was one that mapped from a space to another such that both spaces had the same dimension.

One final question, I also made the argument that a polynomial acted on by the following operator "loses" a dimension: \displaystyle P: A(t) \rightarrow \frac{A(t)-A(0)}{t}
Is that incorrect?

 

[trambolin]

if A(t)-A(0) is a constant then you are out of the space again...

 

[Zorba]

A(t)-A(0) is never constant...

\displaystyle \frac{A(t)-A(0)}{t}=\frac{a_0+a_1t+a_2t^2+a_3t^3+a_4t^4 - a_0}{t}=a_1+a_2t+a_3t^2+a_4t^3

 

[trambolin]

Oops... I don't know what i was thinking. That is correct.