
#1
Feb1813, 03:09 PM

P: 192

Linear operator A is defined as
[tex]A(C_1f(x)+C_2g(x))=C_1Af(x)+C_2Ag(x)[/tex] Question. Is A=5 a linear operator? I know that this is just number but it satisfy relation [tex]5(C_1f(x)+C_2g(x))=C_15f(x)+C_25g(x)[/tex] but it is also scalar. Is function ##A=x## linear operator? It also satisfy [tex]x(C_1f(x)+C_2g(x))=C_1xf(x)+C_2xg(x)[/tex] Thanks for the answer! 



#2
Feb1813, 03:16 PM

Sci Advisor
P: 1,168

I can't understand what you're doing; I would think the operator A takes every function
f(x) into the number 5. I think if you clearly specify your domain and codomain, you will see things more clearly. 



#3
Feb1813, 03:24 PM

P: 192

I define A as multiplicative operator clearly.




#4
Feb1813, 03:54 PM

P: 22

Linear operator
x → cx, where c is a constant, is a linear map.




#5
Feb1813, 04:14 PM

Emeritus
Sci Advisor
PF Gold
P: 8,997

[tex]A(C_1f+C_2g)=C_1Af+C_2Ag.[/tex] f and g are functions. f(x) and g(x) are elements of their codomains. What the equality means is that for all x, [tex](A(C_1f+C_2g))(x)=(C_1Af+C_2Ag)(x)=C_1(Af)(x)+C_2(Ag)(x).[/tex] Edit 2: OK, I see now that the A you had in mind was something different. Suppose that V is some vector space over ℝ, whose elements are functions with a common domain D. I'll assume that D=ℝ. Define ##A:V\to V## by ##Af=5f## for all f. You can easily show that this A is linear using the same method as in my other edit below. Yes, the identity map on ℝ is a linear operator on ℝ. The identity map on any vector space is a linear operator on that vector space. Edit: I see now that that's not the A you had in mind. I stopped reading at "A=x", and assumed that you were denoting the identity map by x. Suppose that V is some vector space over ℝ, whose elements are functions with a common domain D. I'll assume that D=ℝ. Define ##A:V\to V## by saying that for all ##f\in V##, ##Af## is the map from V into V defined by ##Af(x)=xf(x)## for all ##x\in\mathbb R##. (Note that A acts on f, not on f(x). I sometimes use the notation (Af)(x) instead of Af(x) to make that clear. This shouldn't be necessary, since A isn't defined to act on the number f(x), but students often fail to see that). Let ##a,b\in\mathbb R## be arbitrary. For A to be linear, we must have $$A(af+bg)=aAf+bAg.$$ To see if this holds, let ##x\in\mathbb R## be arbitrary. We have $$A(af+bg)(x) = x(af+bg)(x) =x(af(x)+bg(x)) =axf(x)+bxg(x) =aAf(x)+bAg(x) =(aAf+bAg)(x).$$ Since x is arbitrary, this implies that ##A(af+bg)=aAf+bAg##. Since a,b are arbitrary, this means that A is linear. 



#6
Feb1813, 04:38 PM

P: 192

Sorry but I think that you didn't read my post. I defined multiplication operator
which goes from ##f## to ##5f##. 


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