Linear Transformation - Complex -> Complex

In summary: The book is saying that since T(5i) = 0 exists, that means T(x+yi) can equal 0, which is not a one-to-one function. Therefore it is not an isomorphism. In summary, the transformation T(x+yi)= x from C to C is not linear or isomorphic because T(5i)= 0 exists, making it not a one-to-one function.
  • #1
succubus
33
0
The problem:

T(x + yi) = x

C -> C (Complex Numbers)

Show that the above is:

Linear
Isomorphic


This is what i have for showing it's linear:

T(x+yi + a + bi) = x + a + i(y + b) => T(x+a) => T(x) + T(a)
T(k(x+yi)) = k(x) + k(yi) = T(kx) = kT(x)

I assume that i(y+b) = 0. Is this the proper assumption?

Therefore it is not isomorphic because of the fact that i(y+b) = 0?

The answer in the back is states it's not isomorphic because 5i = 0 exists.
So I'm sort of confused as to what they are trying to say.
Aare they saying that by choosing a point on the plane, (x,y) that the component of 5i = 0 (I'm assuming 5 is random) exists for this particular transformation, and since that's impossible the transformation is not isomorphic? Or am I totally on the wrong track.
 
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  • #2
succubus said:
The problem:

T(x + yi) = x

C -> C (Complex Numbers)

Show that the above is:

Linear
Isomorphic


This is what i have for showing it's linear:

T(x+yi + a + bi) = x + a + i(y + b) => T(x+a) => T(x) + T(a)
Completely wrong. T(x+yi + a + bi) is NOT equal to x + a + i(y + b), it is equal to x+ a. And I can make no sense of "=> T(x)+ T(a)".

T(k(x+yi)) = k(x) + k(yi) = T(kx) = kT(x)
NO, T(kx+ yi)= T(kx+ kyi)= kx, not kx+ k(yi). And T(kx)= kT(x) is what you want to prove.


I assume that i(y+b) = 0. Is this the proper assumption?
No, it is not. Where did you get that idea?

Therefore it is not isomorphic because of the fact that i(y+b) = 0?

The answer in the back is states it's not isomorphic because 5i = 0 exists.
So I'm sort of confused as to what they are trying to say.
Probably the problem is that your book doesn't say anything of the sort! Surely you know that "5i" is NOT equal to 0. What your book says is that T(5i)= 0. That is true because T(5i)= T(0+ 5i)= 0. Also T(0)= 0, T(3i)= T(0+ 3i)= 0, etc. Since a an isomorphism must be one-to-one, T is not an isomorphism.

Aare they saying that by choosing a point on the plane, (x,y) that the component of 5i = 0
That makes no sense at all. I have no idea what you mean by "the component of 5i".

(I'm assuming 5 is random) exists for this particular transformation, and since that's impossible the transformation is not isomorphic? Or am I totally on the wrong track.
 
  • #3
Wow, now I feel like a moron.
ha ha ha
Grr...

I got confused when I saw that the transformation went from x + yi to x. :/

Sorry, you're right, the book did say T(5i) = 0 exists, therefore it's not an isomorphism.

I think I get it now
 

FAQ: Linear Transformation - Complex -> Complex

1. What is a linear transformation in complex numbers?

A linear transformation in complex numbers is a function that maps a complex number to another complex number, while preserving the properties of linearity. This means that the transformation must satisfy two conditions: scalar multiplication and vector addition.

2. How is a linear transformation represented in complex numbers?

A linear transformation in complex numbers can be represented as a matrix, where the input vector is multiplied by the matrix to produce the output vector. The matrix is called the transformation matrix and it represents the coefficients of the transformation function.

3. What is the difference between a linear transformation and a non-linear transformation?

A linear transformation preserves the properties of linearity, such as scalar multiplication and vector addition, while a non-linear transformation does not. In other words, a linear transformation can be represented by a matrix, while a non-linear transformation cannot.

4. How can linear transformations be applied in real-world scenarios?

Linear transformations in complex numbers can be applied in various fields, including physics, engineering, and computer graphics. For example, in physics, linear transformations can be used to describe the motion of objects in space, while in computer graphics, they can be used to manipulate images and create 3D graphics.

5. Can a linear transformation change the shape of a complex number?

No, a linear transformation cannot change the shape of a complex number. It can only rotate, reflect, or scale the complex number without changing its overall shape. This is because a linear transformation must preserve the properties of linearity, which includes preserving the shape of the input vector.

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