Undergrad Erroneously finding discrepancy in transpose rule

[nomadreid]

TL;DR

The rule for switching rows and columns to form the transpose of a matrix seems to come up with two different results for a+bi, considered as a scalar or as a matrix. What is my error?

Obviously, there is something elementary I am missing here.

To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix

in the real plane; taking the transpose we get

which then corresponds to a-bi back in the complex plane.

I am making some elementary error here. What?

 

[fresh_42]

$\mathbb{C}\not\cong\mathbb{R}^2$

The complex numbers are a field; the complex plane is not a field. If you want to identify them, or, as in this case, with a subalgebra of $\mathbb{R}^4,$ you have to decide which properties you want to preserve in this identification. Transposition here becomes complex conjugation.

 

[nomadreid]

Interesting! Thank you very much, fresh_42. I will look further into that.

 

[martinbn]

TL;DR: The rule for switching rows and columns to form the transpose of a matrix seems to come up with two different results for a+bi, considered as a scalar or as a matrix. What is my error?

Obviously, there is something elementary I am missing here.

To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix
View attachment 367772
in the real plane; taking the transpose we get
View attachment 367773
which then corresponds to a-bi back in the complex plane.

I am making some elementary error here. What?

What do you mean by a matrix in the real plane? The real plane is not the set of such matrices. More importantly why do you expect this isomorphism (between the complex numbers and the set of such matrices) to respect transpose?

 

[nomadreid]

Thanks very much, martinbn. As fresh_42 pointed out, I was (erroneously) thinking of the real plane along with extra structure, i.e., perhaps as a part of ℝ⁴, and also as fresh_42 pointed out, there isn't the isomorphism which secondary school introductions to the complex plane give the impression of existing, and any correspondence is going to lose some information (such as transpose).

 

[martinbn]

Thanks very much, martinbn. As fresh_42 pointed out, I was (erroneously) thinking of the real plane along with extra structure, i.e., perhaps as a part of ℝ⁴, and also as fresh_42 pointed out, there isn't the isomorphism which secondary school introductions to the complex plane give the impression of existing, and any correspondence is going to lose some information (such as transpose).

But the usual identification of the complex numbers and the real plane sends $a+bi$ to $(a,b)$. You are looking at something different.

 

[nomadreid]

Again, thanks, martinbn. Indeed. I think you mean I oversimplified or neglected the structures that accompany each.

 

[Gavran]

TL;DR: The rule for switching rows and columns to form the transpose of a matrix seems to come up with two different results for a+bi, considered as a scalar or as a matrix. What is my error?

Obviously, there is something elementary I am missing here.

To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix
View attachment 367772
in the real plane; taking the transpose we get
View attachment 367773
which then corresponds to a-bi back in the complex plane.

I am making some elementary error here. What?

There is an isomorphism between the field of complex numbers and a specific set of 2x2 real matrices where a complex number $a+bi$ is mapped to the 2x2 matrix $\begin{pmatrix}a&-b\\b&a\end{pmatrix}$. The conjugation of a complex number corresponds to the transposition of its 2x2 matrix representation.

If you are confused about real numbers, express a real number $a$ as $a+0i$, find its 2x2 matrix representation, and you will see that the matrix is the same as its transposition, as well as the complex number $a+0i$ is the same as its conjugate.

 

[nomadreid]

Thanks, Gavron. These simple examples are very nice.

 

[nomadreid]

The conjugation of a complex number corresponds to the transposition of its 2x2 matrix representation.

So, for example, if we take the Hermitian adjoint (the composition of the complex conjugate and the transpose), I would deduce from your sentence that this would mean that the conjugation of the conjugate of scalar s corresponds to the Hermitian adjoint of its matrix representation; that is, s corresponds to its Hermitian adjoint. But in general the Hermitian adjoint of a scalar equals its complex conjugate. So I am wondering if your choice of the word "isomorphism" for this correspondence was the best one.

 

[Gavran]

So, for example, if we take the Hermitian adjoint (the composition of the complex conjugate and the transpose), I would deduce from your sentence that this would mean that the conjugation of the conjugate of scalar s corresponds to the Hermitian adjoint of its matrix representation; that is, s corresponds to its Hermitian adjoint. But in general the Hermitian adjoint of a scalar equals its complex conjugate. So I am wondering if your choice of the word "isomorphism" for this correspondence was the best one.

The conjugation of a complex number is also a complex number. So the conjugation of a complex number twice corresponds to the transposition of its 2×2 matrix representation twice.

 

[nomadreid]

Thanks, Gavran, but apparently I did not express myself very well. Let me try again. First, if I understand your post correctly (but probably I am not), it seems that in Post #8, that you are stating

Then, however, suppose we take the complex conjugate of both sides , then we get

where α^† is the Hermitian adjoint of α.
Alternatively, suppose we take the transpose of the same both sides, then we get

The problem is that in general

This discrepancy leads me to agree with the earlier answers that one should be careful with the word "isomorphism" here. The correspondence you propose in post #8 does not respect transpose, so the above equations with ≅ would not be correct.

("Would" because I am probably misunderstanding your post. I am happy to be corrected.)

 

[Gavran]

This discrepancy leads me to agree with the earlier answers that one should be careful with the word "isomorphism" here. The correspondence you propose in post #8 does not respect transpose, so the above equations with ≅ would not be correct.

("Would" because I am probably misunderstanding your post. I am happy to be corrected.)

The transposition of a complex number is the same complex number, and the conjugation of a real matrix is the same real matrix. It is obvious that the transposition of a complex number corresponds to the conjugation of its 2×2 matrix representation.

$\alpha^T\cong\begin{pmatrix}a&-b\\b&a\end{pmatrix}^T$ at the beginning of the second line above does not hold, and $\alpha^T$ is not equal to $\bar{\alpha}$.

 

[nomadreid]

Thanks very much, Gavron! I see my error now. I had not rid myself of my prejudice that isomorphism preserved all structures; I understand now that isomorphism only preserves the specific operations that define the structure in question, but that "transpose" is an additional structure. This helps a lot!