Introduction
The notion of matrix determinant is an elementary one which finds its earliest relevance to quantifying changes in volumes of domains under affine transformations. Formally, given an matrix
, the determinant of
is defined as
where denotes the set of
permutations of
letters written in a straight line and
denotes the sign of a permutation
(where we note that
returns
or
only). Now, suppose the entries of
are real-valued measurable functions defined over a common domain
and they are members of the Lebesgue space
. Then, define the integral of the matrix
through the map
whereby
We may denote simply by
because performs integration entry-wise on the matrix
. In this way, linearity of the integral is maintained, and we can give meaning to an expressions such as
by employing the usual partial ordering of square matrices through the notion of positive definiteness (called Loewner order; see Loewner order Wikipedia page). As such, we say that provided the matrix
is positive semi-definite, which is to say
Since maps square matrices to square matrices, we can construct the following notion of determinant integral of a square matrix
.
Definition 1. (Determinant Integral) We say the of an
matrix
with real-valued entries in
is the quantity given by
where is the integral of
.
Example 1. One can show that the following determinant integral of the given matrix returns whenever
and
. I produced this result on New Years Eve 2019:
Commutativity
A natural question to ask is: when can we interchange the integral and determinant without changing the result? In other words, what are the most general conditions on the matrix which guarantee that
Case 
Commutativity trivially holds since
General Set Up for 
In the previous case, our assumption of the entries being in suffices for the integral
to converge. However, this may not be sufficient in general since an arbitrary product of functions in
may not be integrable over
. Therefore, what conditions on the entries of
ensure that each of the
products
are in
? We may stipulate that entries in the
-th row are in
, for some appropriate
with
and
for
. In this way, we require entries are in the same Lebesgue space row-wise. We then have by
inequality
for each permutation , ensuring that the integral
converges and is finite. But then, we also need to ensure that converges so that the determinant integral of
is finite. Under the assumption that
(
) with some exponents
for which
and
for all
, we could assume
is bounded to conclude that each entry of
is in
, by
inequality. However, we would like to keep the domain as arbitrary as possible. So, instead we will assume the following condition:
(H1): The entries of the matrix
are real-valued measurable functions
with
, where
are given exponents such that
and
for all
.
Denote the set of matrices satisfying (H1) by , where
is the vector storing some given exponents that are compatible with (H1).
Closure of
Under Matrix Inversion
As it will soon become an important issue, let us look at the following question:
For an invertible matrix , is its inverse also in
?
To start, consider the simplest case. Suppose we have a matrix :
defined a.e in
that is invertible a.e with inverse
a.e in . Note that The entries of
are all in
. Further, the inverse of
is in
provided the following expressions are all in
:
(1) ……
Assuming were arbitrary, and
a.e in
, we see that
and
are both in
if and only if
. Thus, in this scenario the diagonal elements of
are necessarily in
if
. On the other hand, we generally don’t know whether or not
is integrable, even though due to the (H1) assumption. Because the determinant of
is nonzero a.e in
, one reasonable assumption to impose on
for our discussion is that zero is not a limit point of the image of
over
, i.e there exists a constant
such that
a.e in . Assuming this, we deduce that
which, by (H1), implies the four expressions in (1) are in . Provided the diagonal entries of
are in
, we find that
for
. Hence we have
, for whichever
is given. In summary we have deduced
Proposition 1. Let be a domain, with
given. Suppose
, where
, is such that
, and
- there exists a constant
for which
a.e in
.
Then, is invertible with inverse
.
Now let’s consider the case Let
for some given
such that
and
. We will assume once more that
a.e in
for some constant
, so
is invertible a.e. With
written as
the inverse of is given by
One can show that, provided for all
, then
This follows by visiting each row of
and checking that every product term is in
regardless of whether
is finite or infinite. As an example, let’s consider the first row of
. Since
are in
, we have
As such, , and likewise
. Hence,
, as Lebesgue spaces are vector spaces. Furthermore, suppose
. Then, clearly
. If
, we already know that
, so there is nothing to check in this case. If
, we have
Hence, in all cases , and likewise
. But
Thus, . By assuming
, together with the basic assumption that
, we see that
as well. In all, the first row of has the properties required by (H1). Essential boundedness of the other entries of
, and the (H1) assumption on
, imply that the remaining rows of
have the properties required by (H1) also. We may summarise our observations in the following proposition.
Proposition 2. Let be a domain, with
, and let
with
and
. Suppose
, is such that
, for all
, and
- there exists a constant
for which
a.e in
.
Then, is invertible with inverse
.
We may thus use the assumption of essential boundedness to guarantee if
. Suppose
is invertible almost everywhere in
. It is well-known that the inverse of
is given by the formula
where the adjugate of is given by
a.e in with
, given in terms of the
-minor
of
. As
is the determinant of the
matrix obtained by deleting the
-th row and
-th column of
, this suggests that care needs to be given to developing conditions on
which ensure
. Just as in the two-dimensional case, we will assume that there exists a constant
such that
a.e in
, to simplify our analysis. Then,
, leaving us to determine conditions on
which guarantee
for
. Let
be arbitrary. Then,
is of the form
where is the
matrix given by
Let’s check that our proposed conditions on imply
for each , which would yield
. Since
for each
, we see that
(2) ……
Now let’s impose that all entries of are essentially bounded over
. Then, all entries of
are essentially bounded over
as well. We now check that
. As we saw in the
case, we show first that
is finite for each to get
. This is clear because
via (2). Next, we check that
is finite for each , whenever
, and that
is essentially bounded when
. Indeed,
is essentially bounded as a finite product of essentially bounded functions. Therefore,
, as required in the case
. For
where the final upper bound is finite due to cases . Hence,
and so
for each
, as
and
were arbitrary. In all, we have deduced sufficient conditions that ensure the inverse of a matrix
is also in
.
Proposition 3. Let be given. Suppose
is a domain, and
with
and
. If
, is such that
, for all
, and
- there exists a constant
for which
a.e in
,
then is invertible with inverse
.
Remark 1. We would find it of interest to produce an example of for which
but not all entries of
are essentially bounded or the determinant isn’t uniformly bounded away from zero. This, we leave as an investigation for the reader as we resume our analysis of the commutativity of integration and determinant for matrices in
Non-invertible Matrices
With the (H1) assumption, let’s now consider the case for non-invertible matrices. As a starting example, the equality that we desire in the case
reads
Rearrange to get
If the matrix is symmetric with identical diagonal entries (a.e in ) we have
(3) ……
Notice that this equation implies
if and only if
To simplify (3) further, we can assume that the entries of have zero mean-value over
, which is to say
Note that interpreting this qualification as vanishing mean-value is suitable only for domains with finite -dimensional Lebesgue measure
, in which case the integrals of
and
vanish. However, if
, we’ll just assume this:
Consequently, (3) reduces to
This can hold provided a.e in
, with symmetry and diagonal equality assumed. In fact, without assuming symmetry or equality along the main diagonal of
, we find easily that taking determinant and performing integration are operations that commute for the following set of matrices for
:
The condition that implies that
has linearly dependent rows or linearly dependent columns necessarily. So, in two dimensions the structure of
is simple.
Example 2. In the case , consider the matrix
defined everywhere in
, except at the origin. Then, clearly
over
. Writing
and
, we deduce that
. Calculating directly, we have
which is finite, and likewise
Furthermore,
We thus conclude for each
.
Returning the to general equality which we desire:
we can consider a slightly more general set of conditions that guarantee commutativity. It suffices to assume det and det
a.e in
. Thus, both
and its integral contain linearly dependent rows or columns (a.e in
). So, we let
Therefore, .
Example 3. For and
, let
for some bounded domain
which we’ll later specify. Provided
a.e in
, either
or its transpose has linearly dependent rows. So, without loss of generality, write
for some real-valued measurable function defined a.e over
for which
. Hence,
and since , either the rows of
, or its transpose, are linearly dependent. For the rest of the discussion, suppose that the rows of
are linearly dependent and let’s see where the analysis takes us.
This means then there exists a real constant such that
and
or
(4) ……
and
(5) ……
With given, we have two equations for two unknowns
and
. Eliminating
, we get
or
(6) ……
where
and
Suppose further that and
are linearly independent in
(which we assume is equipped with the standard real inner product). We observe that if
is in the orthogonal complement of
within
, then (6) is satisfied and equations (4), (5) imply that
and
each have vanishing mean-value over
. Consequently,
is reduced to the trivial matrix, implying the unknown
in equations (4), (5) can be any real number. On the other hand, we can construct an example such that
does not annihilate
or
, with
. With such
we can see from (4) that
whereas is determined by (6). However, this solution is possibly non-unique (outside a set of measure zero) because we can add to
any nontrivial member
from the orthogonal complement of
in
to give another solution to (6). Such
exists as
is an infinite-dimensional Hilbert space while
is a finite dimensional subspace. This would suggest that to entertain a general discussion on uniqueness in
, we ought to consider it within
. Such a discussion will be avoided for the moment.
Now, for the example. Let ,
, and
in
. Then, we clearly have
and
Thus (6) is satisfied and
So, the following matrix is a non-trivial member of :
Invertible Matrices
In the context of invertible matrices, we will analyse commutativity of integration and determinant for orthogonal matrices and triangular matrices.
Orthogonal Matrices
Let be a matrix with entries being real-valued measurable functions defined a.e in
. We say
is an orthogonal matrix over
provided
, and
both hold a.e in
. Therefore, taking determinant across either orthogonality equation, we find
is
or
a.e in
, which is to say there exists a set
such that
and
or
for given
. In this case,
is invertible a.e in
with inverse given by its transpose
. If the subset
of
where
assumes the value
has zero
-dimensional Lebesgue measure, we say that
is a special orthogonal matrix over
. Let’s denote the set of orthogonal matrices and special orthogonal matrices defined a.e in
by
and
, respectively. For our discussion, we say that a matrix
is in
if
is a real-valued constant matrix that is orthogonal. Further, if
and
, we say
is in
as a special orthogonal matrix.
Observe that, when , the simplest solution to the equation
(7) ……
is the identity matrix which satisfies
and thus
Note that in general the above equation implies is a constant
almost everywhere in
, which implies either
if
or
is any non-negative number if
. Now suppose
is such that
and
solves (7). Then, we find
So, if we search for such solutions to (7), the underlying domain necessarily has unit volume. Thus, define the set
What we just showed implies then that no element of can satisfy (7) whenever
, but in contrast all elements of
solve (7) when
. In another case, suppose
solves (7) but
. Then, we find
Of course, is non-negative, so we find necessarily
and so
. Therefore, considering the set
we find that if there exists a solution to (7) then
- the domain
has unit volume,
, and
- every element of
solves (7).
Notice that since
. We’ll soon investigate whether or not this inclusion is strict for arbitrary domains
in
. It will also be interesting to determine when either set is empty by virtue of choice of domain and matrix dimension
.
Let’s now consider the following two sets:
Clearly . Now suppose
is given. Then,
is an orthogonal matrix and so there exist sets
such that
,
and
for all
while
for all
. If
solves (7), we find
On the other hand, if solves (7), it follows that
which is equivalent to By the above, provided
satisfies (7) with
then
and so
. Moreover,
. We have thus shown the following. Given a domain
if there exists a solution
to (7) for which
, where
and
are as defined above, then
- the volume of
is strictly larger than 1,
and
- there possibly exists another element
of
that does not satisfy (7).
In fact, if instead satisfies (7) and
, we still have
although
In any case, the volume of the domain must be strictly larger than unity if we seek to establish the existence of a solution
to equation (7), as opposed to assuming unit volume when considering existence over
. All considered, we wonder how the sets
, and
may be characterised.
Triangular Matrices
Let be a matrix of real-valued measurable functions defined almost everywhere in a domain
. If
(for some appropriate
) is a triangular matrix that is a solution to equation (7) we arrive at
(7′) ……
Notice that none of the off-diagonal entries of feature in this equation, but only its diagonal entries appear. Therefore, without loss of generality, it suffices to assume
is a diagonal matrix in our search for solutions to equation (7′). We will also assume
since since the above equation trivially holds when
. Now, assume
is a domain that is radially symmetric with respect to the origin in
(so
and
). Given
, suppose the set of diagonal entries of
contains an odd number of almost everywhere odd functions while the remaining entries are almost everywhere even functions. It follows that, for some
,
is an odd function which satisfies
and so
At the same time, the function
is almost everywhere odd over As such,
In all, equation (7′) holds under these assumptions. But notice that, while our argument shows in this case , we may not necessarily have
a.e in
. Moreover, we can easily produce examples of solutions to equation (7) that indeed lie outside the sets
and
encountered in our discussion for non-invertible matrices.
Example 4. Let be the open unit ball centred at the origin. Define
and
over
. Then, clearly
is radially symmetric with respect to the origin, and
in
while
in
It follows that
while
Then, taking we have an example of an almost everywhere non-degenerate solution in
to equation (7′) for any
,
and
for which
and
for
.
An Integral-Determinant Equation
In this section we study one consequence of commutativity of integration and determinant for invertible matrices. When is a matrix whose determinant integral doesn’t vanish, we know that
is invertible. As such, if we also have
(8) ……
then
Provided satisfies
(9) ……
and
(10) ……
we find
(11) ……
But, we know that
a.e in . This implies equation (11) takes the form
(12) ……
with a.e in
. Going forward, we will first study some basic features associated with equation (12). Then, to simplify our discussion on when equations (8), (9) and (10) all hold, we will consider two kinds of matrices, namely
- triangular matrices, and
- orthogonal matrices.
Later on, we will study some interesting examples of functions and domains for which equation (11) holds. It is there that we generate classes of examples with which beautiful loci can be associated.
Basic Properties
Firstly, we find that the domain is necessarily bounded by unity whenever satisfying equation (12) is in
and positive.
Proposition 4. Let be such that
and let
be a domain. Suppose
is a measurable function in
that is positive a.e in
and satisfies
Then,
, and
if
a.e in
.
Proof. First of all, observe that if , equation (12) cannot hold since then
Thus, . Next, for item 1 we find by Cauchy-Schwarz’s inequality
as required.
For item 2, suppose the contrary holds, that is,
We already know that , so we must have
. However, we find
giving a contradiction. //
Example 5. We find
In particular, there holds
In the upcoming result we unravel what conditions on the domain are required when a function satisfying (12) is allowed to assume negative values. Given
is a measurable function over a domain
, define its positive and negative parts via
and
, respectively, yielding
a.e in
. Suppose
and
are positive everywhere in
and
, respectively, in such a way that
and
, while
and
. After employing a similar argument involving Cauchy-Schwarz inequality and elementary estimation of integrals, we obtain (somewhat) a generalisation of Proposition 4 in
Theorem 1. Let be real numbers for which
and
. Assume
is a domain which we partition into three subsets
such that
,
,
,
, and
Suppose
is a measurable function over
such that
and
are positive over
and
, respectively, and
while
.
If
then
,
, and
where
Moreover, if over
while
over
, then
and
Triangular Matrices
Suppose is upper triangular with
Further, assume satisfies the conditions of Proposition 1, which indicates
is invertible with
also. The desired equation (8) reads
Provided has unit volume and
a.e in
, we see that
while
and so (8) is satisfied. Continuing with this case, we investigate (9) which we would like to be satisfied by
and
Then
with inverse
The integral of is given by
From (9) we necessarily have
(13) ……
and
We are then left with (10) which is equivalent to, in the current case,
This holds without having to make any further assumptions on the matrix . Moreover, notice that (13) is actually (11) in disguise since
is upper-triangular. If
a.e in
also, we have the following elementary proposition in which
is not identity.
Proposition 4. Let be a domain of unit volume, with
given, and let
be such that
and
. Suppose
, given by
a.e in is in
. Then,
and
Let’s return to the desired equation (8), namely
(14) ……
and assume it holds. Assuming satisfies the requirements of Proposition 1, there exists a constant
such that
a.e in
. Hence, neither
nor
can vanish on a subset of
that has positive
-dimensional Lebesgue measure. As such, we calculate
as
a.e in . We again seek to write out the formula given by (9). To this end, we have
with inverse given by
or
due to (14), while
Hence, if and only if
(15) ……
(16) ……
and
(17) ……
Finally, (10) written out reads
(18) ……
Then, by the argument given earlier on, we deduce that
(19) ……
which is precisely
But we observe that with (14), (15), (17), and (18) only, we can derive (19). There is no need for (16) to hold and so there’s no need to satisfy . This leads us to the following result.
Proposition 5. Let be a domain, with
, and let
be such that
and
. Suppose
is given by
a.e in , and satisfies
, and
- there exists a constant
for which
a.e in
.
If there further hold
then
Remark 2. Since no conditions are placed on the entries off the main diagonal of (apart from those due to membership of
in
), we deduce this result also holds in the case of lower triangular
matrices. Moreover, we’d like to think of the equality conditions given in the result as pseudo-intuitive relationships, each of which we believe are starting points for interesting numerical analysis. Furthermore, if we allow for inequality, one can ask how easy it is to produce continuous functions which satisfy
with a given tolerance . We could also ask functional analytic questions on the above inequality. For example, in what function space does the above inequality describe a closed set?
What can be said about upper triangular matrices? If the matrix
is invertible, its inverse is given by
or rather
In this case, (8) reads
(20) ……
Equation (10) for reads
(21) ……
Now,
and if is invertible its inverse is
On the other hand, the integral of is given by
Therefore, (9) implies the following system of equations.
(22) ……
and
Again, we see that some equations aren’t necessary to conclude the relationship we desire. Multiplying (20), (21), and noting equations (22), we deduce
Thus, the equations gotten by equating the non-diagonal entries of (9) aren’t necessary for deriving (11). These calculations, together with Proposition 3, suggest the following generalisation of Proposition 5 to upper-triangular matrices .
Theorem 1. Let be given. Suppose
is a domain, and
is such that
with
. Let
be an upper triangular matrix in
that satisfies
for all
, and
- there exists a constant
for which
a.e in
.
If
and
for , then
Orthogonal Matrices
Let’s consider the relation
(23) …..
in the context of orthogonal matrices . Earlier on we described these matrices when we studied the commutativity of integration and determinant (so recall the set
is where
and the set
is where
for an orthogonal matrix
). Suppose
be a matrix with entries being real-valued measurable functions defined a.e in
.
Provided , the relation
(24) ……
which holds a.e in implies
a.e in . Therefore, if (23) holds for such
,
must have unit volume.
Proposition 6. Let be given and suppose
is a domain of unit volume. If
, then
Observe that whenever , (24) implies
and
assume the same sign a.e in
. Assuming
satisfies (23), we find
if and only if
which is equivalent to
or
(25) ……
since . One may interpret (25) as a condition which says, for a given domain
and a matrix
, there exist two disjoint subsets
of
such that
- the union of
exhausts
in measure,
- the determinant of
is
,
on
respectively, and
- the volumetric difference between
and
is unity.
So, above we could take and
, for example. Moreover, when (25) holds, it suggests
is -1 more often than +1 if
, or +1 more often than -1 which
would indicate. In all, we can state
Theorem 2. Let be given and assume
is a domain. Suppose
is such that there exist sets
for which
- the union of
exhausts
with respect to
-dimensional Lebesgue measure,
- the determinant of
is
,
on
respectively, and
- the volumetric difference between
and
is unity.
Then
Copyright © 2021 Yohance Osborne