linear algebra « Bedtime mathematics

matrix commutators

December 26, 2009 mymbl Leave a comment

We know that set of commutators of ${n\times n}$ matrices is exactly the space of matrices of trace zero and has dimension ${n^2-1}$ . I discuss an interesting question related to this which was brought to my notice by Anamika:

Given a matrix ${C}$ of trace zero can we construct two matrices ${A}$ and ${B}$ of determinant ${1}$ , such that ${AB - BA = C}$ . Here is a proposed way of finding matrices ${A}$ and ${B}$ such that ${AB - BA = C}$ .

It turns out that this problem has a variety of applications and it goes by the name Sylvester equation (after fixing one matrix say A). For example, it is used in Computer Vision to estimate the pose(position, orientation ) of an object based on a series of images. See video here

We first try a simple case: Suppose ${D}$ is matrix with non-zero entries only in the diagonal and at locations ${i,(i+1)}$ for ${i=1,2,\ldots, n-1}$ (like in the Jordan Normal form) with ${[D]_{i\overline{i+1}} = 1}$ or ${0}$ . We want to find a matrix ${X}$ such that ${DX - XD = C}$ where ${C}$ is a ${n\times n}$ matrix with trace zero. We equate the ${i,j}$ entries on both sides to get,

$\displaystyle [C]_{ij} = \sum_{k=1}^n ( [D]_{ik} [X]_{kj} - [X]_{ik} [D]_{kj})$

expanding we get

$\displaystyle [C]_{ij} = [D]_{ii} [X]_{ij} - [X]_{ij} [D]_{jj} + [D]_{i \overline{i+1}}[X]_{\overline{i+1}j} - [X]_{i\overline{j-1}}[D]_{\overline{j-1}j}$

Now if ${\text{vec}(X)}$ denotes the operator that flattens the rectangular(column-wise) matrix to a column vector then we need to solve the system.

$\displaystyle \breve{A} \cdot \text{vec}(X) = \text{vec}(C)$

where,

We can solve the system ${\breve{A}\,\text{vec}(X) = \text{vec}(C)}$ provided ${\breve{A}}$ has maximum rank. We can reconstruct ${X}$ from ${\text{vec}(X)}$ but this does not ensure that the matrix ${X}$ is of determinant ${1}$ . We can ensure that ${X}$ has determinant ${1}$ by addition of a suitable scalar times identity matrix ( which must also be a solution trivially)

Wikipedia mentions that there is a better way to represent this matrix using tensor product ${\otimes}$ ( also called Kronecker product of matrices http://en.wikipedia.org/wiki/Kronecker_product):

$\displaystyle \breve{A} = I_n \otimes A - A^T \otimes I_n$

where ${I_n}$ is the ${n\times n}$ identity matrix.

0.1. Related results

On a related note, here is another question. Can you produce matrices ${E}$ and ${F}$ such that the commutator ${EF - FE}$ is the sum of two other ${n \times n}$ commutators ${AB-BA}$ and ${CD-DC}$ ? . Using the methods above we can find ${E}$ and ${F}$ indirectly, but is there a simple representation of ${E}$ and ${F}$ in terms of ${A,B,C,D}$ ? I do not know. However here are some insights:

Suppose ${e_{ij}}$ denotes the matrix with ${1}$ only at the position ${i,j}$ , ( ${1\leq i,j \leq n}$ ), in the matrix and the rest filled by zeros. We use the fact that the set of ${n\times n}$ matrices which are commutators is a linear space with basis:

$\displaystyle e_{ij} = e_{ii}e_{ij} - e_{ij}e_{ji} \text{ for } i \neq j$

and

$\displaystyle e_{ii} - e_{i+1\,i+1}$

I will leave this to the diligent reader to see why this is true. This says that any ${n\times n}$ commutator ${CD - DC}$ is a linear combination of these basis vectors. Armed with this result we can break the original problem down to the following two problems:

Given matrices ${n\times n}$ matrices ${A,B}$ with commutator ${AB-BA}$ , the commutator ${e_{ij}}$ ( ${j \neq i}$ ) and a scalar ${c}$ , find ${E}$ and ${F}$ such that
$\displaystyle AB - BA + c (e_{ij}) = EF -FE$

.
Similarly given a scalar ${c}$ , and matrices ${A,B}$ , for each ${i}$ , find matrices ${E,F}$ such that
$\displaystyle AB - BA + c( e_{ii} - e_{i+1\,i+1}) = EF - FE$

I could not give a simple representation of ${E,F}$ in terms of ${A,B,e_{ij}}$ . For for some special cases, it seems possible. Suppose you know that for the index ${j}$ , ${B_j}$ is the only column of ${B}$ containing non-zero entries. Now choose constants ${c_k = \text{signum } b_{kj}}$ , ${k=1,2\ldots,n}$ , such that ${\sum_k c_k b_{kj} \neq 0}$ .

We next note the identity ${e_{ij} = e_{ik}e_{kj} - e_{kj}e_{ij}}$ , for ${i \neq j}$ and ${1\leq k\leq n}$ , using which we can write:

$\displaystyle (\sum_k |b_{kj}|) e_{ij} = \sum_k (c_k e_{ik}) (b_{kj} e_{kj}) - {b_{kj}e_{kj}}c_{k}e_{ik}$

where ${B_{kj}}$ denotes the entry at the location ${k,j}$ in the matrix ${B}$ . Now let ${B_j = \sum_k B_{kj}e_{kj}}$ . Next using the fact that ${e_{ik}e_{lj} = 0}$ is ${k \neq l}$ , we have ${e_{ik} B_{kj}e_{kj} = e_{ij}B_j}$ and ${B_j e_{ik} = 0 = \sum_k B_{kj}e_{kj}}$ , so the above expression for ${e_{ij}}$ can be rewritten as: