Showing that multiplication by a complex number is a linear transform

[Eclair_de_XII]

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.

 

[Gaussian97]

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.

 

[PeroK]

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.

 

[HallsofIvy]

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).

 

[Gaussian97]

Yes, but given a basis for the initial and final space, any (finite) linear transformation can be written as a matrix.

 

[WWGD]

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!

 

[WWGD]

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.

 

[WWGD]

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.

 

[FactChecker]

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.