# Linear operator

1. Feb 18, 2013

### matematikuvol

Linear operator A is defined as
$$A(C_1f(x)+C_2g(x))=C_1Af(x)+C_2Ag(x)$$
Question. Is A=5 a linear operator? I know that this is just number but it satisfy relation
$$5(C_1f(x)+C_2g(x))=C_15f(x)+C_25g(x)$$
but it is also scalar.
Is function $A=x$ linear operator? It also satisfy
$$x(C_1f(x)+C_2g(x))=C_1xf(x)+C_2xg(x)$$

2. Feb 18, 2013

### Bacle2

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. Feb 18, 2013

### matematikuvol

I define A as multiplicative operator clearly.

4. Feb 18, 2013

### Ferramentarius

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

5. Feb 18, 2013

### Fredrik

Staff Emeritus
This should be
$$A(C_1f+C_2g)=C_1Af+C_2Ag.$$ f and g are functions. f(x) and g(x) are elements of their codomains. What the equality means is that for all x,
$$(A(C_1f+C_2g))(x)=(C_1Af+C_2Ag)(x)=C_1(Af)(x)+C_2(Ag)(x).$$
The number 5 isn't, but the map $x\mapsto 5$ is. For every real number t, there's a "constant function" $C_t:\mathbb R\to\mathbb R$ defined by $C_t(x)=t$ for all $x\in\mathbb R$. These functions are linear operators on ℝ.

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.

I wouldn't write that definition like that. x is a variable (that typically represents a real number, not a function). You want to define A as the map $x\mapsto x$. This is called "the identity map". It's sometimes denoted by I or id, and shouldn't be denoted by x. The proper way to define A as the identity map without using those notations is to say this: Define $A:\mathbb R\to\mathbb R$ by A(x)=x for all $x\in\mathbb R$.

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.

Last edited: Feb 18, 2013
6. Feb 18, 2013

### matematikuvol

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

7. Feb 18, 2013

### Fredrik

Staff Emeritus
OK, I'll edit that part too.