MHB Rotations, Complex Matrices and Real Matrices

[Math Amateur]

I am reading Kristopher Tapp's book: Matrix Groups for Undergraduates.

I am currently focussed on and studying Section 1 in Chapter2, namely:

"1. Complex Matrices as Real Matrices".I need help in fully understanding what Tapp is saying in this section regarding the function

$$ \rho_n \ : \ M_n ( \mathbb{C} ) \rightarrow M_{2n} ( \mathbb{R} ) $$Section 1, Chapter2 reads as follows:
View attachment 3990
View attachment 3991I am having trouble fully understanding how the function:

$$ \rho_n \ : \ M_n ( \mathbb{C} ) \rightarrow M_{2n} ( \mathbb{R} $$

relates to $$f_n$$ and $$R_{ \rho_n (A) }$$ ...

For example if

$$ \rho_2 \begin{pmatrix} a+bi & c+di \\ e + fi & h+ji \end{pmatrix}
= \begin{pmatrix} a & b & c & d \\ -b & a & -d & c \\ e & f & h & j \\ -f & e & -j & h \end{pmatrix}$$
My question is what is

$$f_n$$ and $$R_{ \rho_n (A) }$$

in this case, and how exactly do these expressions relate to $$\rho_n$$ ... ...?
Hope someone can help?
Now in the above the linear transformation $$R_A$$ is mentioned ... ... $$R_A$$ is defined in Tapp Ch. 1, Section 5 as follows:
https://www.physicsforums.com/attachments/3992
View attachment 3993

 

[Fallen Angel]

Hi Peter,

I don't know if I have understood your issue.

For example, when $n=2$

$f_{2}:\Bbb{C}^2\longrightarrow \Bbb{R}^4$ is given by

$f_{2}(a+bi,c+di)=(a,b,c,d)$

And $R_{\rho_{2}(A)}$ is the multiplication of a row vector times this $4\times 4$ matrix you have typed on.

The relation $\rho_{n}$ gives you is how you can define the same application modulo the canonical isomorphism from $\Bbb{C}^{n}$ to $\Bbb{R}^{n}$ , looking at it like a complex function or a real one.

 

[Euge]

I am reading Kristopher Tapp's book: Matrix Groups for Undergraduates.

I am currently focussed on and studying Section 1 in Chapter2, namely:

"1. Complex Matrices as Real Matrices".I need help in fully understanding what Tapp is saying in this section regarding the function

$$ \rho_n \ : \ M_n ( \mathbb{C} ) \rightarrow M_{2n} ( \mathbb{R} ) $$Section 1, Chapter2 reads as follows:

I am having trouble fully understanding how the function:

$$ \rho_n \ : \ M_n ( \mathbb{C} ) \rightarrow M_{2n} ( \mathbb{R} $$

relates to $$f_n$$ and $$R_{ \rho_n (A) }$$ ...

For example if

$$ \rho_2 \begin{pmatrix} a+bi & c+di \\ e + fi & h+ji \end{pmatrix}
= \begin{pmatrix} a & b & c & d \\ -b & a & -d & c \\ e & f & h & j \\ -f & e & -j & h \end{pmatrix}$$
My question is what is

$$f_n$$ and $$R_{ \rho_n (A) }$$

in this case, and how exactly do these expressions relate to $$\rho_n$$ ... ...?
Hope someone can help?
Now in the above the linear transformation $$R_A$$ is mentioned ... ... $$R_A$$ is defined in Tapp Ch. 1, Section 5 as follows:

Peter,

First, let me say that there is something confusing going on in the first section you posted. As written, the matrix $B$ represents clockwise rotation through angle $\theta$, not counterclockwise rotation. It can't be a difference in convention because the computation $R_B(r\cos \phi, r\sin \phi) = (r\cos(\theta + \phi), r\sin(\theta + \phi))$ suggests that the intended matrix is

$$B = \begin{pmatrix}\cos \theta & -\sin \theta\\\sin \theta & \cos \theta\end{pmatrix}$$

However, after the computation of $R_A(e^{i\theta})$, the clockwise convention is used! So I'll be using the clockwise convention in this discussion.

Given a matrix $A\in M_n(\Bbb C)$, $\rho_n(A)\in M_{2n}(\Bbb R)$. So $A$ represents the linear map $R_A : \Bbb C^n \to \Bbb C^n$ and $\rho_n(A)$ represents the linear map $R_{\rho_n(A)} : \Bbb R^{2n} \to \Bbb R^{2n}$. That is, $R_A(X) = XA$ for all $X\in M_n(\Bbb C)$ and $R_{\rho_n(A)}(Y) = Y\rho_n(A)$ for all $Y\in M_{2n}(\Bbb R)$. We want to relate $R_A$ and $R_{\rho_n(A)}$ by a linear map from $\Bbb C^n \to \Bbb R^{2n}$, because then we can determine $R_{\rho_n(A)}$, and hence $\rho_n(A)$. This is where $f_n$ comes in. To have $R_{\rho_n(A)} \circ f_n = f_n \circ R_A$ means that

$$R_{\rho_n}(A)(f_n(c_1 + \mathbf{i}d_1,\cdots, c_n + \mathbf{i}d_n)) = f_n(R_A(c_1 + \mathbf{i}d_1,\ldots, c_n + \mathbf{i}d_n))$$

that is, if $R_A(c_1 + \mathbf{i}d_1,\ldots, c_n + \mathbf{i}d_n) = (x_1 + \mathbf{i}y_1,\ldots, x_n + \mathbf{i}y_n)$, then

$$R_{\rho_n}(A)(c_1,d_1,\ldots,c_n,d_n) = (x_1,y_1,\ldots, x_n,y_n).$$

Unraveling this a bit further, we find that if

$$A = \begin{pmatrix}a_{11} + \mathbf{i}b_{11}&\cdots&a_{1n} + \mathbf{i}b_{1n}\\a_{21} + \mathbf{i}b_{21}&\cdots&a_{2n} + \mathbf{i}b_{nn}\\ \vdots & & \vdots\\a_{n1} + \mathbf{i}b_{n1}&\cdots &a_{nn} + \mathbf{i}b_{nn}\end{pmatrix}$$

then for $n = 1$,

$$\rho_n(A) = \begin{pmatrix}a_{11} & b_{11}\\-b_{11} & a_{11}\end{pmatrix}$$

and for $n > 1$,

$$\rho_n(A) = \begin{pmatrix}\rho_1(a_{11} + \mathbf{i}b_{11})&\cdots &\rho_1(a_{1n} + \mathbf{i}b_{1n})\\ \rho_1(a_{21} + \mathbf{i}b_{21})&\cdots&\rho_1(a_{2n} + \mathbf{i}b_{2n})\\ \vdots & & \vdots\\ \rho_1(a_{n1} + \mathbf{i}b_{n1}) & \cdots & \rho_1(a_{nn} + \mathbf{i}b_{nn})\end{pmatrix}$$

As you can see here, I have written the expression of $\rho(A)$ in block form, where each block is determined by the image of an entry in $A$ under $\rho_1$.

 

[Fallen Angel]

Hi Euge,

Matrix $B$ is clockwise if you work with a column vector $v$ and then compute $Bv$, but I understand he is working with row vectors $w$, so in this case $wB$ is a counterclockwise rotation. It is not usual to work with row vectors and postmultiplicate the matrix, but it's still correct.

 

[Euge]

Yes, Fallen Angel, that's right. I was going to edit that but thanks for adding that comment.

 

[Math Amateur]

Yes, Fallen Angel, that's right. I was going to edit that but thanks for adding that comment.

Thanks to the posts of Euge and Fallen Angel ... and some further work and reflection, I now have a much better understanding of Tapp's section on "Complex matrices as real matrices" ... ... i was really confused there for a while ... so many thanks to Euge and Fallen Angle for their help ...Just a further issue however ... in the initial case where:

$$\rho_1(a+bi) = \begin{pmatrix} a & b \\ -b & a \end{pmatrix}$$

it is clear from the geometry and Tapp's analysis that:

$$a = \cos \theta$$ and $$b = \sin \theta$$... ... BUT ... what is the situation with respect to $$a,b,c, d, e, f, h, j$$ in $$ \rho_2 \begin{pmatrix} a+bi & c+di \\ e + fi & h+ji \end{pmatrix}
= \begin{pmatrix} a & b & c & d \\ -b & a & -d & c \\ e & f & h & j \\ -f & e & -j & h \end{pmatrix}$$Hope someone can help?

Peter

 

[Fallen Angel]

Hi Peter,

In that case, you got the same situation BUT you are working in a 2-dimensional complex vector space, so you have 4 rotation angles, each one corresponding to the rotation of one of the two entry complex numbers with respect of one of the two complex planes , but this can't be drawn or easily visualized.

The cosinus and sinus of this angles are $(a,b), (c,d), (e,f)$ and $ (g,h)$.

 

[Math Amateur]

Hi Peter,

In that case, you got the same situation BUT you are working in a 2-dimensional complex vector space, so you have 4 rotation angles, each one corresponding to the rotation of one of the two entry complex numbers with respect of one of the two complex planes , but this can't be drawn or easily visualized.

The cosinus and sinus of this angles are $(a,b), (c,d), (e,f)$ and $ (g,h)$.

Thanks for your further help Fallen Angel ... I obviously need to learn more about complex vector spaces ...

Thanks again,

Peter