Linear operators & dimension

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

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

Now if we consider the subspace of these polynomials such that [tex]a_0=0,\ a_1=0,\ a_2=0}[/tex], I propose that the dimension of of this subspace is 2 (versus the dimension of [tex]P_4[/tex] 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 [tex]P_n[/tex]
 

HallsofIvy

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

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

Now if we consider the subspace of these polynomials such that [tex]a_0=0,\ a_1=0,\ a_2=0}[/tex], I propose that the dimension of of this subspace is 2 (versus the dimension of [tex]P_4[/tex] which is 5. Am I incorrect in saying this?
Yes, any such polynomial can be written as [itex]0(1)+ 0(x)+ 0(x^2)+ a_3 t^3+ a_4t^4[/itex] and so [itex]\{t^3, t^4\}[/itex] is a basis.

Based on the answer to this I have a follow up question regarding a linear operator on [tex]P_n[/tex]
Fire away!
 
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(Cheers)

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

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

He seemed to argue that since [tex]tA(t)[/tex] now belongs to [tex]P_5[/tex] it now has an extra dimension, but I argue that since [tex]tA(t)[/tex] has the form [tex]a_0t+a_1t^2+a_2t^3+a_3t^4+a_4t^5[/tex] and since a polynomial in [tex]P_5[/tex] would be [tex]A(t)=a_0+a_1t^1+a_2t^2+a_3t^3+a_4t^4+a_5t^5[/tex] that it does indeed have the same dimension... :uhh:
 
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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).
 
77
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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: [tex]\displaystyle P: A(t) \rightarrow \frac{A(t)-A(0)}{t}[/tex]
Is that incorrect?
 
if A(t)-A(0) is a constant then you are out of the space again...
 
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[tex]A(t)-A(0)[/tex] is never constant...

[tex]\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[/tex]
 
Oops... I don't know what i was thinking. That is correct.
 

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