What is the transpose of a matrix $i[A,B]$?

[Slimy0233]

Homework Statement

Q.6. If A and B are Hermitian matrices, then which of the following matrices will
be neither Hermitian nor skew hermitian in nature?

Relevant Equations

For a Hermitian Matrix A,
1. $A^{\dagger} = A$
2. $iA$ is a skew-Hermitian Matrix

I don't know how to simplify option b)

I mean, I don't know how to take the transpose of a matrix which exists as a product of $\iota$, how can I solve this further?

I need help only with option c. I have attempted this previously and have gotten what seems to be the right answer, but I don't know how I did that now

 

[PeroK]

In general your solutions have a step missing. It's not a major problem, but you leave a bit of work for the reader.

You also need to assume that $A, B \ne 0$.

 

[Slimy0233]

In general your solutions have a step missing. It's not a major problem, but you leave a bit of work for the reader.

You also need to assume that $A, B \ne 0$.

except for opt B and opt D you don't need extra steps as you can use property
The sum of two Hermitian Matrices is also a Hermitian matrix. But, I still get how to solve c though. I want to take the transpose of i[A,B], how can I do this?

 

[PeroK]

I want to take the transpose of i[A,B], how can I do this?

I thought you ready did in your OP.

 

[Slimy0233]

I thought you ready did in your OP.

I didn't unfortunately, I did try and do it now which is attached, but I am pretty sure there's an easier way as I am not supposed to do this in the exams.

 

[PeroK]

That's okay. Perhaps it could be simpler, but all the steps are correct.

 

[PeroK]

For example, you could first show that $[A,B]$ is skew-Hermitian, hence $i[A, B]$ is Hermitian.

 

[Slimy0233]

For example, you could first show that $[A,B]$ is skew-Hermitian, hence $i[A, B]$ is Hermitian.

Thank you! of c! I forgot the fact that [A, B] can also be treated like a matrix (even though it doesn't look like a conventional matrix)
This is what I was looking for

 

[Slimy0233]

For example, you could first show that $[A,B]$ is skew-Hermitian, hence $i[A, B]$ is Hermitian.

if A is skew-Hermitian then $(iA)^{\dagger} = iA$ making $iA$ skew-Hermitian$$(iA)^\dagger = i^*A^\dagger = -iA^\dagger = -i (-A) = iA$$
edit: I corrected something stupid I said

 

[vanhees71]

You meant if $A$ is skew-Hermitean than $\mathrm{i} A$ is Hermitean, and that's in fact what you correctly have proven.

 

[WWGD]

If i may ask what the i stands for in
i[a,b]? I guess [a,b] is the conmutator?

 

[Mark44]

If i may ask what the i stands for in
i[a,b]? I guess [a,b] is the conmutator?

I'd be willing to bet it's the imaginary unit...

 

[WWGD]

I'd be willing to bet it's the imaginary unit...

So you're multiplying the conmutator by the imaginary unit?

 

[Mark44]

So you're multiplying the conmutator by the imaginary unit?

Commutators are a topic I've never studied, but from the work shown, i is multiplying the result of the commutator action.

 

[WWGD]

Commutators are a topic I've never studied, but from the work shown, i is multiplying the result of the commutator action.

I _am_ multiplying. ;).

 

[Mark44]

I _am_ multiplying. ;).

You may be, but so is $i$.

 

[PeroK]

Commutators are a topic I've never studied, but from the work shown, i is multiplying the result of the commutator action.

It's standard stuff, as far as QM is concerned.

 

[WWGD]

It's standard stuff, as far as QM is concerned.

Which is above my level at this point.