Linear operators & dimension

1. Dec 14, 2009

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 $$P_n$$

2. Dec 14, 2009

HallsofIvy

Staff Emeritus
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.

Fire away!

Last edited: Dec 15, 2009
3. Dec 14, 2009

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... :uhh:

Last edited: Dec 14, 2009
4. Dec 14, 2009

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).

5. Dec 14, 2009

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?

6. Dec 14, 2009

trambolin

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

7. Dec 14, 2009

Zorba

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

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

8. Dec 14, 2009

trambolin

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