Short Communication - Journal of Applied Mathematics and Statistical Applications (2019) Volume 2, Issue 1

## More on the orthogonal complement functions

**Robert Jennrich**

^{1*}and Albert Satorr^{2}^{1}Department of Mathematics, University of California, Los Angeles, USA

^{2}Department of Mathematics, Universitat Pompeu Fabra, Barcelona, Spain

- *Corresponding Author:
- Robert Jennrich

Department of Mathematics,University of California

Los Angeles, USA

**Tel:**(310) 825-2207

**E-mail:**[email protected]

**Accepted Date:** February 23, 2019

**Citation: **Jennrich R, Satorr A. More on the orthogonal complement functions. J Appl Math Statist Appl. 2019;2(1):47-50.

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### Abstract

Continuous orthogonal complement functions have had an interesting history in covariance structure analysis. They were used in a seminal paper by Browne in his development of a distribution-free goodness of fit test for an arbitrary covariance structure. The proof of his main result Proposition 4 used a locally continuous orthogonal complement function, but because he failed to show such functions existed his proof was incomplete. In spite of the fact that his test had been used extensively, this problem was not noticed until 2013 when Jennrich and Satorra pointed out that his proof was incomplete and completed it by showing that locally continuous orthogonal complement functions exist. This was done using the implicit function theorem. A problem with the implicit function approach is that it does not give a formula for the locally continuous function produced. This problem was potentially solved by Browne and Shapiro who gave a very simple formula F(X) for an orthogonal complement of X. Unfortunately, they failed to prove that their function actually produced orthogonal complements. We will prove that given a p×q matrix X0 with full column rank qKeywords

Covariance structure analysis, Distribution free tests, Implicit function theorem, QR factorization.

## Introduction

The orthogonal complement of a *p×q* matrix X with *q<p* and full
column rank is a *p*×(*p-q*) matrix *Y *such that [X,Y] is invertible.
Note that *Y* must have full column rank. Jennrich and Satorra in
Theorem 1 show how to compute an orthogonal complement *Y*
of an arbitrary *p×q* matrix *X* with full column rank *q<p* using
the long form of a QR factorization. Unfortunately, *Y* is not a
continuous function of X [1].

In a seminal paper Browne showed how to test the goodness of fit of an arbitrary covariance structure [2]. The proof of his main result Proposition 4 used a locally continuous orthogonal complement function, but because he failed to show such functions existed his proof was incomplete. In spite of the fact that his test had been used extensively this problem was not noticed until 2013 when Jennrich and Satorra pointed out that his proof was incomplete and completed it by showing that locally continuous orthogonal complement functions exist [1]. This was done using the implicit function theorem. A problem with the implicit function approach is that it does not give a formula for the function produced and Jennrich and Satorra conjectured their function could not be expressed by an explicit formula.

Browne and Shapiro say that this conjecture is incorrect by presenting an explicit formula for a locally continuous orthogonal complement function [3]. Browne and Shapiro in

slightly different notation state the following: Let *X*_{0} be any *p×q*
matrix with full column rank *q*<*p* and X^{c}_{0} be any orthogonal complement of *X*_{0}. Consider the following matrix valued
function:

Let N be any neighbourhood of *X*_{0} such that *X* has full column
rank for all *X* ϵ *N*. Browne and Shapiro say that on *N, F(X)*
is a well-defined continuous function and claim *F(X)* is an
orthogonal complement of *X* for all *X* in *N*. They, however, fail
to show this. In particular they fail to show *F(X)* has full column
rank. As a consequence their claim is only a conjecture. We will
begin by proving their conjecture.

**Proof of the Browne-Shapiro conjecture**

**Lemma 1:** Let *X*_{0} be a *p*×*q* matrix of full column rank *q*<*p*.
There is a neighbourhood *N* of *X*_{0} that contains only full column
rank matrices.

**Proof:** Let X be an arbitrary p×q matrix and let ɡ(*X*) =det(*X*'*X* ) .
Since ɡ(*X* ) is continuous there is *δ*>0 such that

It follows from the last inequality that ɡ( *X* ) ≠ 0 and hence that
det(*X*'*X* ) ≠ 0 and hence that *X* has full column rank for all *X* in
the neighbourhood

**Lemma 2:** Let *X*_{0} be a *p*×*q* matrix with full column rank *q*<*p*
and let *X*^{c}_{0} be any orthogonal complement of *X*_{0}. Then

is continuous and well defined for *X* ∈ *N* .

**Proof: **By Lemma 1 all *X* ∈ *N* have full column rank. Thus
F(*X*) is well defined. It follows from the continuity of matrix
multiplication and matrix inversion that *F*(*X*) is continuous for
all *X* ∈ *N*.

**Theorem 1:** Let *X*_{0} be a *p*×*q* matrix with full column rank *q*<*p*.
Let *X*^{c}_{0} be any orthogonal complement of *X*_{0}. Then there is a
neighbourhood N of *X*_{0} such that *X* has full column rank for all
*X* ∈ *N* and

is an orthogonal complement of *X*.

**Proof:** Note that *F*(*X*) is *p*×(*p*-*q*) and *F*(*X*) is orthogonal
to *X*. It is sufficient to prove *F*(*X*) has full column rank. Let
ɡ(*X* ) =det(*F*(*X* )′*F*(*X*) ) . Then ɡ( *X* ) is a continuous and it
follows from this that there is *δ*>0 such that

It follows from the last inequality that ɡ( *X* ) ≠ 0 . Hence
det(*F*(*X* )′*F*(*X*) )≠ 0 and hence *F*(*X*) has full column rank.

This proof is much simpler than that given by Jennrich and
Satorra using the implicit function theorem [1]. In their paper
Browne and Shapiro prove Browne's Proposition 4 without
using their *F*(*X*) formula. One could, however, use Theorem
1 because it asserts the existence of a locally continuous
orthogonal complement function and this is all that is needed
to fix Browne's proof of Proposition 4. There are now three
proofs of Browne's Proposition 4, the one given by Jennrich and
Satorra [1], the one given by Browne and Shapiro [3], and the
one using Theorem 1.

A problem with using Theorem 1 is that for an *X* of interest one
has no way of knowing if *X* is in *N* since the only thing we know
about *δ* is that it is greater than zero. If necessary for an *X* of
interest one can always compute an orthogonal complement of
*X* using Theorem 1 of Jennrich and Satorra, but it will not be a
continuous function of *X *[1].

See the following example which shows numerically that
Theorem 1 of Jennrich and Satorra produces orthogonal
complements as does Theorem 1 of this document and also
suggests an interesting alternative method using *F*(*X*) [1].

**Theory:**

**Example:**

Let

Then *X* is a *p* by *q* matrix with *p* = 4 and *q* = 2. Theorem 1 of
Jennrich and Satorra [1] says that if

*X*= *QR*

is the long form of a QR factorization of X, then the last p-q columns of Q are an orthogonal complement of X. Computing a QR factorization of X gives

The last *p*-*q* columns of this are:

and this is an orthogonal complement *X*. This demonstrates
numerically that Theorem 1 of Jennrich and Satorra [1] produces
orthogonal complements.

Note that when *X*=*X*_{0}, *X* satisfies the assumptions of Theorem 1
and the computed value of

which is an orthogonal complement of *X*. Thus in this case at
least *F*(*X*) is an orthogonal complement of *X* as asserted by
Theorem 1.

Let

be a random matrix whose components are independent standard normal variables. Note that it has full column rank. The value of the Browne-Shapiro.

This has full column rank and hence is an orthogonal complement
of *X* even though *X* may not satisfy the basic assumption of
Theorem 1 above because *X* is simply a random matrix.

When this is repeated 1000 times in every case *F*(*X*) is an
orthogonal complement of *X*. This suggests that *F*(*X*) is an
orthogonal complement of *X* with high probability.

We show below that much more is true. We show *F*(*X*) fails
to be an orthogonal complement of *X* only for *X* in subset of
Labesgue measure zero in *R ^{pxq}*.

Note that when *X* = *X*^{c}_{0} , F(*X*)=0 and hence *F*(*X*) is not an
orthogonal complement of X for all X with full column rank.

## A necessary and sufficient condition for F(X) to be an orthogonal complement of X.

To simplify the development that follows let us begin by looking
at the simplest possible case when *p* = 2 and *q* = 1. Assume *X* is not in the column space of
*X*^{c}_{0} . Then in **Figure 1** clearly, *F*(*X*)
is a non-zero vector and has full column rank. Thus *F*(*X*) is an
orthogonal complement of *X* for all *X* not in the column space
*X*^{c}_{0} . If* X* is in the column space of
*X*^{c}_{0} , then *F*(*X*)=0 and hence
*F*(*X*) is not an orthogonal complement of *X*. Note this happens
only on a set that has Labesgue measure zero in *R*^{2}. We will
show that this happens for arbitrary *p* by *q*<*p* matrices *X*.

We begin with the following theorem,

**Theorem 2:** Let *F*(*X*) be the Browne-Shapiro formula. That
[*X*,*X*^{c}_{0}] is invertible is a necessary and sufficient condition for
*F*(*X*) to be an orthogonal complement of *X*.

**Proof:** Assume [*X*,* X*^{c}_{0}] has full column rank. Assume *F*(*X*)
does not have full column rank. Then

*F*(*X*)a=0

for some vector a *a* ≠ 0. Thus

It follows that* X*^{c}_{0}a is in the column space of *X*. Thus

*X*^{c}_{0}a = Xb

for some vector b. Thus there is a vector in the column space
of *X* that is in the column space of
*X*^{c}_{0} . It follows that [*X*,* X*^{c}_{0}]
does not have full column rank. This contradiction proves *F*(*X*)
has full column rank when [*X*,* X*^{c}_{0}] has full column rank. This is
a sufficient condition for *F*(*X*) to be an orthogonal complement
of *X*.

If [*X*,* X*^{c}_{0}] does not have full column rank, then
Xa = *X*^{c}_{0}b for two vectors a and b not both zero. If *a* = 0, then *b* = 0. This
implies *a* ≠ 0 which in turn implies *b* ≠ 0. Now

which implies that *F*(*X*) does not have full column rank. Thus
*F*(*X*) is not an orthogonal complement of *X* when [*X*,* X*^{c}_{0}]
does not have full column rank. Thus *F*(*X*) is an orthogonal
complement of *X* if and only if [*X*,* X*^{c}_{0}] has full column rank.

## How often does F(X) fail to be an orthogonal complement of X?

**Lemma 2:** A polynomial in *n* variables is either identically 0 or
its roots have Labesgue measure zero in *R ^{n}*.

**Proof:** The proof is by mathematical induction on *n*. For* n*=1,
*P*(*x*) of degree *d* can have at most d roots, which gives us the base
case. Now, assume that the theorem holds for all polynomials in
*n*-1 variables. Let and write

Let *I*[*A*] be the indicator function of a set *A*. By Fubini's theorem,

The first two integral equalities follow from Fubini's Theorem.
The last integral equality follows from the fact that given
is a polynomial in *n*-1 variables and by the induction
hypothesis this is identically zero for all or its zeros have
Labesgue measure zero in R^{n-1}[4].

**Comment:** Lemma 2 is a simple very interesting result which
apparently is not in the literature. It is not in any journals covered
in Jstor. It is not in any of the five analysis books we own and
according to Caron and Traynor it is not in any measure theory
book.

**Theorem 3:** The *X* for which the Browne - Shapiro function
*F(X)* is not an orthogonal complement of *X* have Labesgue
measure zero in *R ^{p×q}*

^{}.

**Proof: **F(X) is not an orthogonal complement of X if and
only if [*X*,* X*^{c}_{0}] is singular. This happens if and only if
det( [*X*,* X*^{c}_{0}][*X*,* X*^{c}_{0}])=0
. As a function of X this is not
identically zero because it is not zero when X=X_{0}. Moreover
det det( [*X*,* X*^{c}_{0}][*X*,* X*^{c}_{0}])
is a polynomial in *p×q *variables *x*=*vec*(*X*)
which is not identically zero. It follows from Lemma 2 that its
zeros have Lebesgue measure zero in Rpq. Or equivalently the
*X* such that *F*(*X*) is not an orthogonal complement of *X* have
Labesgue measure zero in *R ^{p×q}*.

**Remark: **Theorem 3 implies that *F(X)* almost never fails to be
an orthogonal complement of *X*. If *X* is a sample from a density,
then *F(X)* is an orthogonal complement of *X* with probability
one.

## Conclusion

Browne and Sharpiro [3] give a formula for computing
orthogonal complements. Let *X*_{0} be any *p*×*q* matrix with full
column rank *q*<*p* and *X*^{c}_{0} be any orthogonal complement of
*X*_{0}. Browne and Sharpiro function is

They state that if *N* be any neighbourhood of *X*_{0} such that *X*
has full column rank for all *X* ϵ *N* then *F*(*X*) is an orthogonal
complement of *X* for all *X* in *N*. They, however, fail to show this.
In particular they fail to show *F*(*X*) has full column rank. As a
consequence their claim is only a conjecture. The main point of
our paper is to prove their conjecture and much more. We prove
that on all but a set of *p*×*q* matrices *X* with Labesque measure
zero, *F*(*X*) is an orthogonal complement of *X*. In particular if *X* is a sample from a density,* F*(*X*) is an orthogonal complement
of *X* with probability one.

Theorem 1 is a proof for the Browne-Sharpiro conjecture. We
also prove additional results that should enhance considerably
the practical relevance of the Browne-Shapiro function for
an orthogonal complement. Theorem 2 gives a necessary and
sufficient condition for *F*(*X*) to be an orthogonal complement of
*X*. Lemma 2 is a not well known, but very useful result about the
roots of an arbitrary polynomial in n variables. If the polynomial
is not identically zero, its roots have Labesque measure zero.
Theorem 3 is our main result.

## Acknowledgements

The authors would like to thank Ellen Jennrich for proof reading our paper and helping with its submission. The research of the second author is supported by grant EC02014-59885-P from the Spanish Ministry of Science and Innovation.

## References

- Jennrich RI, Satorra A. Continuous orthogonal complement functions and distribution free goodness of fit tests for covariance structure analysis. Psychometrika. 2013;78: 445-552.
- Browne MW. Asymptotically distribution-free methods for the analysis of covariance structures. Psychometrika. 1984;37: 62-83.
- Browne MW, Shapiro A. A comment on the asymptotics of a distribution free goodness of fit test statistic. Psychometrika. 2013;78: 196-9.
- Caron R, Traynor T. The zero set of a polynomial. 2005.