Linear Transformations (or lack thereof)

[nateHI]

Homework Statement

Let V be the set of complex numbers regarded as a vector space over the real numbers R. Find a linear transformation T: V → V which is not complex linear (i.e. not a linear transformation if V is regarded as a vector space over the complex numbers).

Homework Equations

V=R(i)

The Attempt at a Solution

I have a solution. What I need is a better explanation of what complex linear means. Or perhaps a rewording of the problem statement so I might have a better understanding of why my solution is a solution.

 

[tiny-tim]

hi nateHI!

What I need is a better explanation of what complex linear means. Or perhaps a rewording of the problem statement so I might have a better understanding of why my solution is a solution.

f is linear if f(λa + µb) = λf(a) + µf(b) for all scalars λ and µ

so what happens eg if λ = i ?

 

[maajdl]

"... regarded as a vector space over the real numbers R"

What does that mean?

 

[dirk_mec1]

I think they mean z = u +iv with (u,v) in RxR.

 

[pasmith]

Homework Statement

Let V be the set of complex numbers regarded as a vector space over the real numbers R. Find a linear transformation T: V → V which is not complex linear (i.e. not a linear transformation if V is regarded as a vector space over the complex numbers).

Homework Equations

V=R(i)

The Attempt at a Solution

I have a solution. What I need is a better explanation of what complex linear means. Or perhaps a rewording of the problem statement so I might have a better understanding of why my solution is a solution.

The complex numbers are a 1-dimensional vector space over themselves, so a complex-linear map $\mathbb{C} \to \mathbb{C}$ is a map of the form $z \mapsto az$ for some $a \in \mathbb{C}$, ie. multiplication by a constant.

The map which takes $z = x + iy$ to its complex conjugate $z^{*} = x - iy$ is not of that form (there is no $a \in \mathbb{C}$ such that $z^{*} = az$ for all $z \in \mathbb{C}$), and so is not complex linear. But it is real linear (it's just $(x,y) \mapsto (x, -y)$).

Therefore the map which takes $z$ to its real part $\frac12 (z + z^{*})$ is not complex linear either.

 

[dirk_mec1]

The map which takes $z = x + iy$ to its complex conjugate $z^{*} = x - iy$ is not of that form (there is no $a \in \mathbb{C}$ such that $z^{*} = az$ for all $z \in \mathbb{C}$), and so is not complex linear. But it is real linear (it's just $(x,y) \mapsto (x, -y)$).

No not true, take z = i then z* = -i, take a =-1 and you got it.

 

[pasmith]

No not true, take z = i then z* = -i, take a =-1 and you got it.

Your $a$ must work FOR ALL $z \in \mathbb{C}$! Finding an $a$ which works for one $z$ is irrelevant!

 

[dirk_mec1]

You're right my bad.

 

[nateHI]

OK, I get it now. Thanks pasmith and everyone else.