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I'm a physics grad student working, for my research, in a matrix vector space over the complex numbers. That is, both operators and the vectors they act upon are $n\times n$ complex matrices living in the same space. When it comes to the eigenvalue equation I've been using

$AB-BA=\lambda B$,

where $A$ and $B$ are $n\times n$ complex matrices. I can "justify" this choice based on the Heisenberg picture of Quantum Mechanics, but I would like to give it a more solid mathematical background. So in this regard I have several questions:

Is the choice above sensible? what other choices could I try? and last what is the relevant literature dealing with matrix vector spaces? I mean, every semi-decent book mentions the space of matrices as a vector space, but that's just about it. I've found some valuable pieces of information scattered througout the problems in the Hoffman linear algebra text, but again, that's just about it. I have yet to find a text that treats matrix vector spaces in any depth. Any help would be really appreciated.

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Can you edit your title to be more informative to the content of the post? – Uticensis Mar 12 '11 at 20:08
up vote 4 down vote accepted

The linear operator $\text{ad}_A : B \mapsto AB - BA = [A, B]$ defines the adjoint representation of the Lie algebra $\mathfrak{gl}_n$, and what you are studying are its eigenvalues. You don't want a linear algebra book; you want a book on Lie theory. The classic reference (although it is not particularly directed towards physicists) is Fulton and Harris's Representation Theory: a First Course.

The important structure here is not the vector space but the action on it, which is why books on linear algebra won't say anything specific about "matrix vector spaces" because they are, after all, still vector spaces.

What other choices you have depends, of course, on what it is you're doing.

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Thanks a lot, I now realize I was looking in the wrong places! – Ricardo Mar 12 '11 at 23:54
I am sure you can find a reference more directed towards physical applications if you ask around your department, also. – Qiaochu Yuan Mar 13 '11 at 11:15
just a nitpick, I don't think the adjoint representation is accurate per se when considering the Heisenberg picture. Taking a finite dimensional analogue, the propagator $\exp iHt$ is an element of the unitary group $U(n)$, and you can treat $H$ then as something living in the Lie algebra $u(n)$. The state space being $n$ dimensional, the observable $A$ should be a self-adjoint map $A\in M^n$. The fact that you can treat $H$ and $A$ both as elements of $gl(n)$ is more a happy accident, I think. So while I agree what the OP want is probably some representation theory of... – Willie Wong Mar 15 '11 at 1:48
...classical groups, I am not sure whether the prescription to focus on the adjoint representation of $gl(n)$ is the right one. – Willie Wong Mar 15 '11 at 1:50
@Willie: sure, but given that the OP explicitly mentions working with this representation I thought I might as well give the term to look up in a book. – Qiaochu Yuan Mar 15 '11 at 1:53

it is good to see you here. First let me start why your problem is not correctly stated. This also explains why my MO answer was correct. One can see an equation by using different glasses, but no one considers $3x=\lambda x$ as an eigenvalue value problem or a PDE! To have an eigenvalue problem you should have a correctly stated linear operator $T$ whose action on a vector space is a priori known. Solving an eigenvalue problem means determining the eigenvalues and the corresponding eigenspaces of that operator.

The main reason why not every matrix equation containing a $\lambda$ is an eigenvalue equation is that not every matrix corresponds to a well defined operator. (remember that a matrix is nothing more than a rectangular representaion of a linear map)

Now, answers to your questions: First of all, I still don't see a correctly stated problem. Is $\lambda$ given and are you looking for $A+B?$ Or, is $A$ given and are you looking for $\lambda$ or $B?$ And most importantly, what do you want to obtain at the end of the day?

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this is a correctly stated eigenvalue equation. The linear operator, as I have said, is $B \mapsto AB - BA$. – Qiaochu Yuan Mar 13 '11 at 11:15
Yes, the operator is $A$ and the vector it acts upon is $B$ ($A$ is Hermitian by the way) and they are both $n\times n$ matrices. Given $A$ and $\lambda$ I want to determine $B$, just like in the ordinary (right) eigenvalue equation $Ax=\lambda x$ where $x$ is a column vector. Now, since you can also define a left eigenvalue equation, I thought the natural choice, in a matrix space, would be to use the anti-commutator $AB+BA=\lambda B$ but, for reasons specific to my research, I needed operators acting from the right to include a minus sign – Ricardo Mar 13 '11 at 21:19
and so I ended with the expression posted in the question and wanted to check if it was (mathematically) well defined, and learn more about matrix vector spaces in the process. Most physicists would regard these questions as nitty picky – Ricardo Mar 13 '11 at 21:32
so I find this site really helpful. Thank you both for your comments – Ricardo Mar 13 '11 at 21:41

Here's something that might help. Using Krnoecker products, we can write

$$AX-XA = 0$$


$$[(I \otimes A) - (A^T \otimes I)] \vec X = 0.$$

This is taken directly from Horn and Johnson's Topics in Matrix Analysis, pg. 257.  

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