FOR MOORE-PENROSE INVERSES IN HILBERT SPACES
DANIEL ST ˘ANIC ˘A
We construct an efficient algorithm (similar to the Gram-Schmidt algorithm) for the computation of the Moore-Penrose inverse of a linear operator with closed range between real Hilbert spaces.
AMS 2000 Subject Classification: 47A05, 65F20.
Key words: Moore-Penrose inverse, Hilbert space, linear operator, closed range.
The role of Moore-Penrose inverse of a linear bounded operator with closed range between Hilbert spaces, in statistics, prediction theory, control system analysis, curve fitting and numerical analysis is well recognized, as many publications show (see, for example, [1]). Consequently, it is important, both practically and theoretically, to find a good algorithm for computing a Moore-Penrose inverse (see, for example, [2], [3], [4], [5], [6]).
In this paper we propose an efficient algorithm (similar to the Gram- Schmidt algorithm for constructing an orthonormal basis in finite Hilbert spaces) for the computation of the Moore-Penrose inverse of a linear bounded operator. In the case of an operator with finite range, this algorithm is direct and implies a number of iterations equal to the dimension of the range. More- over, for an operator whose domain has an orthonormal countable basis, the Moore-Penrose inverse is obtained as limit of operators with finite range.
LetXandY be two real Hilbert spaces and letL(X, Y) denote the space of all bounded linear operators betweenX andY. Consider T ∈ L(X, Y). Let R(T),N(T),T∗ denote the range, the null space, and the adjoint ofT and let LC(X, Y) denote the space of all bounded linear operators with closed range.
Let T ∈ LC(X, Y) and let P (respectively, Q) denote the orthogonal projection onR(T∗) (respectively onR(T)). Then there exists a unique linear bounded operator T+:Y →X satisfying
(1) T+T =P, T T+=Q and T+T T+=T+.
MATH. REPORTS11(61),2 (2009), 165–170
We callT+ the Moore-Penrose inverse ofT. Giveny∈Y, it is well know that x=T+yis the unique elementxinA:=n
x0∈X| kT x0−yk= inf
z∈XkT z−yko such that kxk ≤ kx0k, for all x0 ∈A.
Suppose that dimR(T∗) =n. Let e1, e2, . . . , en be a basis in R(T∗) and fi = T ei for all i = 1, n. If u ∈ X and v ∈ Y, we denote by uv∗ the linear operator defined as
uv∗ :Y →X, uv∗y=hv, yiu for all y∈Y.
We define the elements (e0i)ni=2,(x)ni=1⊂X, (fi0)ni=2,(yi)ni=1⊂Y,by x1= 1
kf1ke1 and y1=T x1, (we shall see further that f1 6= 0) and
e0i =ei−
i−1
X
j=1
hfi, yjixj, fi0 =T e0i and xi= 1
kfi0ke0i, yi =T xi
for i= 2, n (we shall see thatfi 6= 0).
We consider the linear operatorU :Y →X defined as
(2) U :=
n
X
i=1
xiy∗i.
We shall prove that U =T+ (Moore-Penrose inverse ofT).
Lemma 1. R(U)⊆R(T∗).
Proof. We proceed by induction. We prove that xi ∈R(T∗), ∀i= 1, n.
For i = 1 the assertion is true from the definition of x1 (since e1 ∈ R(T∗) = N(T)⊥,we deduce that f1 6= 0). Assume thati= 1, k (k≤n−1)fi0 6= 0 and xi∈R(T∗).We then have
e0k+1 =ek+1−
k
X
j=1
hfk+1, yjixj ∈R(T∗) =N(T)⊥
(from the hypothesis). Therefore, fk+10 =T e0k+16= 0 and xk+1= 1
kfk+10 ke0k+1∈R(T∗).
Letx∈R(U). Then there exists y∈Y such that x=U y=
n
X
i=1
hyi, yixi ∈R(T∗).
Hence R(U)⊆R(T∗).
Lemma 2. N(T∗)⊆N(U).
Proof. Letx∈N(T∗). Then U x=
n
X
i=1
hyi, xixi=
n
X
i=1
hxi, T∗xixi= 0.
Consequently, N(T∗)⊆N(U).
Proposition 1. The elements y1, y2, . . . , yn are an orthonormal basis in R(T).
Proof. Obviously,kyik= 1, ∀i= 1, n.
We have hy2, y1i= 1
kf20khf2− hf2, y1iy1, y1i= 1 kf20k
hf2, y1i − hf2, y1i ky1k2
= 0.
Let k≤n−1.Suppose thathyi, y1i= 0,for all i= 1, k. Then hyk+1, y1i= 1
kfk+10 k
fk+1−
k
X
i=1
hfk+1, yiiyi, y1
=
= 1
kfk+10 k hfk+1, y1i − hfk+1, y1i ky1k2
= 0.
Therefore, hyi, y1i= 0,for all i= 2, n.
Let i ∈ 2, n and k < i−1. Suppose that hyi, yji = 0, for all j = 1, k.
Then
hyi, yk+1i= 1 kfi0k
fi−
i−1
X
j=1
hfi, yjiyj, yk+1
=
= 1
kfi0k
hfi, yk+1i − hfi, yk+1i kyk+1k2
= 0.
Therefore, hyi, yji= 0,for all i, j= 1, n cu i6=j.
Proposition 2. The elementsx1, x2, . . . , xn are a basis inR(T∗).
Proof. If we assume that there existsi∈1, nsuch thatxi= 0,thenyi= 0,which is a contradiction. Therefore,xi6= 0,∀i= 1, n.Letα1, α2, . . . , αn∈R such that α1x1+α2x2+· · ·+αnxn = 0.Then α1y1+α2y2+· · ·+αnyn= 0 and, by Proposition 1, we have αi = 0, ∀i = 1, n. To conclude, the vectors x1, x2, . . . , xn are a basis inR(T∗).
Theorem 1. Let X and Y be two Hilbert spaces and T ∈ LC(X, Y).
Assume thatdimR(T∗) =n. Then the operatorU defined by(2)is the Moore- Penrose inverse of T (U =T+).
Proof. Let x ∈ X. We have X = R(T∗) +R(T∗)⊥ = R(T∗) +N(T).
There is a unique decomposition x := w+x1 with w := P x ∈ R(T∗) and x1 ∈ N(T). It follows from Proposition 2 that there exists real numbers α1, α2, . . . , αn such thatw=α1x1+α2x2+· · ·+αnxn. Then
U T x=U T w=U T n
X
i=1
αixi
=
n
X
i=1
αiU yi =
n
X
i=1
αixi =w=P x .
Therefore, U T =P.
Since xi = U yi, ∀i= 1, n,the vectors x1, x2, . . . , xn belong to R(U), so that dimR(U)≥n. Since from Lemma 1,R(U)⊆R(T∗) and dimR(T∗) =n, we have R(U) =R(T∗).
It follows from Lemma 2 that N(U) ⊇ N(T∗) and N(U)⊥ ⊆ N(T∗)⊥. Then R(U∗)⊆R(T). Since
dimR(U∗) = dimR(U) = dimR(T∗) = dimR(T) =n,
we have R(U∗) =R(T). BecauseY =N(T∗) +R(T) =N(U) +R(U∗), from the previously equality we obtain N(T∗) =N(U).
Let y ∈ Y. Then we have a unique decomposition y := u+y1 with u := Qy ∈ R(T) ¸si y1 ∈ N(T∗). By Proposition 1 there exist real numbers β1, β2, . . . , βn such thatu=β1y1+β2y2+· · ·+βnyn. Then
T U y=T U u+T U y1 =T U u=T U n
X
i=1
βiyi
=
n
X
i=1
βiyi =u=Qy.
Therefore, T U =Q. We have
U T U y=U T U u=U u=U y .
So, U T U =U and U verifies equations (1). It follows that U =T+. Let beXandY two real Hilbert spaces. Assume thatXhas a countable orthogonal basis (xn)n∈N. Denote byXn the subspace ofX generated by the set {xi |i∈1, n}, putYn=T(Xn) and denote by Qn the projection operator on Fn. In [2, Theorem 6] is proved the results below.
Theorem 2. Let T ∈ LC(X, Y). Then lim
n→∞Tn+y=T+y for ally ∈Y. Then, forn∈Nwe construct the operatorTn(which has a finite dimen- sional range) and, using the previous algorithm, we determine the operator denoted by Un := Tn+. It follows from Theorem 2 that T+y = lim
n→∞Uny for all y∈Y.
Example1. Let
T =
1 0 −1
0 1 0
1 1 −1 1 2 −1
with dimR(T∗) = 2. Choosinge1 =
1 0
−1
,e2 =
0 1 0
(a basis inR(T∗)), we obtain
f1 =
2 0 2 2
, f2 =
0 1 1 2
.
Then
x1 = 1 2√
3
1 0
−1
, y1= 1
√3
1 0 1 1
.
Further,
e02 =e2− hf2, y1ix1 = 1 2
−1 2 1
and f20 =T e02 =
−1 1 0 1
.
Then
x2 = 1 2√
3
−1 2 1
, y2 = 1
√3
−1 1 0 1
.
It follows that
T+=x1y∗1+x2y2∗= 1 6
2 −1 1 0
−2 2 0 2
−2 1 −1 0
.
Example 2. Let us solve the linear least squares problem: find c ∈ R4 such that
kAc−bk= inf
x∈R4
kAx−bk,
where
A:=
1 1 2 0
1 2 1 2
2 1 6 −3 0 1 2 −1
1 0 1 0
and b:=
1 2 3 4 5
.
Choosing e1 =
1 1 2 0
, e2 =
1 2 1 2
and e3 =
2 1 6
−3
(a basis in
R(T∗), we obtain
A+=
0.08928571 0.02678571 0.09821428 −0.49107143 0.43750000 0.01785714 0.20535714 −0.08035714 0.40178571 −0.31250000 0.03571428 0.03574128 0.08928571 0.05357143 5.34E−20 0.07142857 0.19642857 −0.07142857 −0.14285714 0.12500000
and
c=
0.66071429 0.23214286 0.58928571 0.30357143
.
REFERENCES
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Wiley-Interscience, New York, 2003.
[2] I. Ichim,A new solution for the least squares problem. Internat. J. Comput. Math.72 (1999), 207–222.
[3] G.R. Luecke,A numerical procedure for computing the Moore-Penrose inverse. Numer.
Math.32(1979),2, 129–137.
[4] N. Shinozaki, M. Sibuya and K. Tanabe, Numerical algorithms for the Moore-Penrose inverse of a matrix: Direct methods. J. Optim. Theory Appl.22(1977),1, 193–203.
[5] D. St˘anic˘a, A new algorithm for the pseudoinverse in Hilbert spaces. Rev. Roumaine Math. Pures Appl.50(2005), 367–372.
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Received 9 December 2008 University of Bucharest
Faculty of Mathematics and Computer Science Str. Academiei 14
010014 Bucharest, Romania
