Question

y(x) : C --> R is not a linear functional because C is a complex vector space. I do not know why. Please explain me. I can see they are not linear functional by checking only. Is there any logic on it? The problem is : y (x) : C --> R defined by y(x) = x1 where x = x1 + ix2 My attempt: let x = 1+i in C; t = 1-i in C, alpha1 = 1; alpha2 = i then y (alpha1 x + alpha2 t) = 2 which is not alpha1 y(x) + alpha2 y(t) = 1+i The conclusion is y is not a linear functional.

Answer 1

@wio

Answer 2

What do you mean by linear function? Linear transformation?

Answer 3

I am not sure about the notation. But it may not that, since i have a linear transformation part, not this part.

Answer 4

can you define linear function?

Answer 5

yes, I have it now

Answer 6

A linear functional on a vector space V is a scalar-value function y defined for every vector x with the property that (identically in the vectors x1 and x2 and the scalars alpha1 and alpha2)
y(alpha1x1 + alpha2 x2) = alpha1y(x1)+ alpha2y(x2)

Answer 7

Well consider...

Answer 8

\(f(ix) = if(x)\)

Answer 9

We know that \(if(x)\in \mathbb C\setminus \mathbb R\) and \(f(ix)\in \mathbb R\).

Answer 10

Unless \(f(x)=0\)

Answer 11

how f(ix) \(\in\) R?

Answer 12

Because \(f:\mathbb C\to \mathbb R\).

Answer 13

you defined f(ix) = i f(x). so, (ix) in C , under the function f, it is = i f(x) --> if(x) is in R but it cannot be, since i \(\in\) C

Answer 14

Unless \(f(x) = 0\), we can say \(f(ix) \neq if(x)\). This breaks the \(f(cx)=cf(x)\) property of linear functions.

Answer 15

oh, you use contradict method?

Answer 16

Yeah

Answer 17

so, in definition, the "scalar-value function" implied that y (x) is a number, right?

Answer 18

Huh?

Answer 19

in example, they said
For x = (\(\xi_1,....., \xi_n\)) in \(C^n\), write y(x) = \(\xi_1\), it's a number.

Answer 20

In more general, if y(x) = linear combination of \(\alpha_i \xi_i\) , it's a number, still.

Answer 21

A 'number'?

Answer 22

yes, sir. Let take x =(1,2) \(\in R^2\), x is a vector, but 1,2 are numbers. and if y is defined by y(x) = x1, then y(x) = 1 is a number, am I right?

Answer 23

or if y is defined by linear combination of x1, x2, let say 2x1 + 3 x2, then y(x) = 2*1 +3*2 = 6, it's a number, too.

Answer 24

Number? Are we talking about a scalar or a complex number?

Answer 25

I lost you. :)
Is there any relationship between "a scalar" and " a complex number"?
to me, a scalar is a number, it can be a real number or a complex number.

Answer 26

depend on the field we compute, if it is / R , then a scalar is a real number. If it is / C , then the scalar is a complex number.

Answer 27

Number is very vague...

Answer 28

Are you saying that \(y:\mathbb C^n\to \mathbb R^n\)?

Answer 29

and \(X^n\) is \(\forall x_i\in X:\langle x_1,x_2,\dots, x_n\rangle \)

Answer 30

C or R is a field with dimension 1, right?

Answer 31

so, we can see its elements be a vector with 1 dimension.

Answer 32

You can think of them that way.

Answer 33

bingo. I meet you again

Answer 34

Then a scalar is a one dimensional vector.

Answer 35

oh yea, I got you now. Let me check the problem on book. I am sorry for my carelessness on typing,

Answer 36

since complex vector space has at least 2 dimensions ( if we consider it is a real vector space)and at most 4 dimensions ( if we consider it is a complex itself), therefore, the function f : C --> R can be considered as f: R^2--> R

Answer 37

I dunno, I sort of gave my own attempt to prove by contradiction. If my proof doesn't make sense, then you need to explain how it contradicts the problem.