Showing that multiplication by a complex number is a linear transform

In summary, the problem is asking for a linear transformation on the real vector space ##\mathbb{R}^2##. If you can show that this is the case, you will have solved the problem.
  • #1
Eclair_de_XII
1,083
91
Homework Statement
"The set ##\mathbb{C}## of complex numbers can be canonically identified with the space ##\mathbb{R}^2## by treating each ##z=x+iy\in \mathbb{C}## as a column ##(x,y)^T\in \mathbb{R}^2##.

(a) "Treating ##\mathbb{C}## as a complex vector space, show that multiplication by ##\alpha=a+ib\in \mathbb{C}## is a linear transformation in ##\mathbb{C}##. What is the matrix?"
Relevant Equations
##(x+iy)(a+ib)=(xa-by)+i(ya+bx)##
If I had to guess what the complex matrix would look like, it would be:

##T(x+iy)=(xa-by)+i(ya+bx)=\begin{pmatrix}
a+bi & 0 \\
0 & -b+ai\end{pmatrix}\begin{pmatrix}
x \\
y \end{pmatrix}##

I'm not too sure on where everything goes; it's my first time fiddling with complex numbers, and moreover, using them in linear algebra.
 
Physics news on Phys.org
  • #2
Mmm... I think there's something wrong. ##\mathbb{C}## is a dimension 1 complex vector space. So of course, multiplication by ##\alpha## is a linear transformation, but it has no associated matrix (well, a matrix 1##\times##1 if you want). So I think the problem wants to show that multiplication by ##\alpha## is a linear transformation in ##\mathbb{R}^2##. In that case, the ##2\times 2## matrix must have only real numbers.
 
  • Like
Likes member 587159
  • #3
Eclair_de_XII said:
Problem Statement: "The set ##\mathbb{C}## of complex numbers can be canonically identified with the space ##\mathbb{R}^2## by treating each ##z=x+iy\in \mathbb{C}## as a column ##(x,y)^T\in \mathbb{R}^2##.

(a) "Treating ##\mathbb{C}## as a complex vector space, show that multiplication by ##\alpha=a+ib\in \mathbb{C}## is a linear transformation in ##\mathbb{C}##. What is the matrix?"
Relevant Equations: ##(x+iy)(a+ib)=(xa-by)+i(ya+bx)##

If I had to guess what the complex matrix would look like, it would be:

##T(x+iy)=(xa-by)+i(ya+bx)=\begin{pmatrix}
a+bi & 0 \\
0 & -b+ai\end{pmatrix}\begin{pmatrix}
x \\
y \end{pmatrix}##

I'm not too sure on where everything goes; it's my first time fiddling with complex numbers, and moreover, using them in linear algebra.

The question perhaps wants you to present multiplication by ##\alpha## as a linear transformation on the real vector space ##\mathbb{R}^2##.

In this case, the set of complex numbers is represented by 2D vectors and also by a set of 2x2 real matrices.
 
  • #4
A linear transformation doesn't have to be written as a matrix. A transformation, T, is linear if and only if T(u+ v)= Tu+ Tv and T(pv)= pT(v).
Let u= x+ iy and v= p+ iq. Then u+ v= (x+ p)+ i(y+ q). T(u+v)= (a+ bi)((x+p)+ i(y+q))= (a(x+ p)- b(y+ q))+ i(b(x+ p)+ a(y+ q))= (ax+ ap- by- bq)+ i(bx+ bq+ ay+ aq). And (a+ ib)(x+ iy)= (ax- by)+ i(ay+ bx) and (a+ ib)(p+ iq)= (ap- bq)+ i(aq+ bp). Adding those (a+ ib)(x+ iy)+ (a+ ib)(p+ iq)= (ax- by+ ap- ba)+ i(ay+ bx+ aq+ bp). Do you see that proves T(u+ v)= Tu+ Tv? (That is actually just saying that "multiplication distributes over addition in the complex numbers".) Similarly, if u= x+ iy then pu= px+ ipy. T(pu)= (a+ ib)(px+ ipy)= apx- bpy+ i(apy+ bpx)= p[ax- by+ i(ay+ bx)]= aT(u).
 
  • #5
Yes, but given a basis for the initial and final space, any (finite) linear transformation can be written as a matrix.
 
  • #6
HallsofIvy said:
A linear transformation doesn't have to be written as a matrix. A transformation, T, is linear if and only if T(u+ v)= Tu+ Tv and T(pv)= pT(v).
Let u= x+ iy and v= p+ iq. Then u+ v= (x+ p)+ i(y+ q). T(u+v)= (a+ bi)((x+p)+ i(y+q))= (a(x+ p)- b(y+ q))+ i(b(x+ p)+ a(y+ q))= (ax+ ap- by- bq)+ i(bx+ bq+ ay+ aq). And (a+ ib)(x+ iy)= (ax- by)+ i(ay+ bx) and (a+ ib)(p+ iq)= (ap- bq)+ i(aq+ bp). Adding those (a+ ib)(x+ iy)+ (a+ ib)(p+ iq)= (ax- by+ ap- ba)+ i(ay+ bx+ aq+ bp). Do you see that proves T(u+ v)= Tu+ Tv? (That is actually just saying that "multiplication distributes over addition in the complex numbers".) Similarly, if u= x+ iy then pu= px+ ipy. T(pu)= (a+ ib)(px+ ipy)= apx- bpy+ i(apy+ bpx)= p[ax- by+ i(ay+ bx)]= aT(u).
Welcome back! Great to see you here again!
 
  • #7
Gaussian97 said:
Yes, but given a basis for the initial and final space, any (finite) linear transformation can be written as a matrix.
To be pretentious, there is an isomorphism between L(V,V) and nxn matrices once an ordered basis has been chosen. I am not sure of what "category" we are talking about, i.e., what kind of objects we're considering L(V,V) and all matrices.
 
  • #8
Ultimately, using polar coordinates, you see that Complex multiplication is a composition of scaling and rotation, each being linear, so the composition is linear. This gives you a basis-free argument.
 
  • #9
Represent u+iv in a matrix form as ## \begin{pmatrix} u & -v \\ v & u \end{pmatrix} ## Multiply two complex numbers as matrices and show that the result represents the desired complex number. Then you should be able to use matrix properties.
 

1. What is a complex number?

A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers and i is the imaginary unit (defined as the square root of -1).

2. What does it mean for multiplication by a complex number to be a linear transform?

A linear transform is a mathematical function that maps one vector space to another in a linear manner. In the case of multiplication by a complex number, it means that the resulting transformation can be represented as a matrix multiplication.

3. How do you show that multiplication by a complex number is a linear transform?

To show that multiplication by a complex number is a linear transform, we must demonstrate that it satisfies the properties of linearity. This includes the properties of additivity and homogeneity.

4. Can you provide an example of multiplication by a complex number as a linear transform?

Yes, consider the complex number 2 + 3i. Multiplication by this complex number can be represented as the matrix [2 3; -3 2] and can be applied to a vector [x y] by multiplying the matrix by the vector to get the transformed vector [2x+3y -3x+2y].

5. Why is it important to understand that multiplication by a complex number is a linear transform?

Understanding that multiplication by a complex number is a linear transform is important because it allows us to use the tools and techniques of linear algebra to analyze and solve problems involving complex numbers. This can be especially useful in fields such as engineering, physics, and computer science.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
464
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
780
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
20
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
798
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
Back
Top