SIAM J. APPL. MATH.
Vol. 24, No. 3, May 1973
GENERALIZED
INVERSION OF MODIFIED
MATRICES*
CARL D. MEYER, JR.t
forthe
Abstract.For an m x n complexmatrixA and two columns,c and d, representations
Moore-Penroseinverseofthe matrixA + cd* are givenforall possiblecases. Moreover,each representationinvolvesonlyA, At,c, d, and theirconjugatetransposes.
For a square,nonsingular
1. Introduction.
matrixA, it is well known
bya matrix
ofrank1 to produceM = A
[2,pp. 173-178]thatwhenA is modified
ofM, ifitexists,
is givenbytheformula
+ cd*,theinverse
M-1 = A`
(1.1)
A- ed*A-',
-
where,B= 1 + d*A-lc. By meansof (1.1), one can alterone or moreof the
matrix[1], [3,p. 79], [5].
ofA and stilluse A- 1 to invertthemodified
elements
or
schemesknownas rankannihilation
Also,(1.1)is thebasisfortheinversion
ifone is dealingwith
thereinforcement
method[6], [2,pp. 173-178].However,
eithera rectangular
matrixor a squaresingularmatrixA, as is oftenthecase,
obtainedtheMoore-Penrose
and he has previously
inverse[4] At ofA, itis not
matrix
alwayspossibleto obtainthe Moore-Penroseinverseof themodified
A + cd* by using(1.1) with(- 1) replacedby (t). Earlierworkby Cline [8],
allowedone to obtainthe Moore-Penroseinverseof A + cd* in the special
the
caseswhenAd*c = 0 or elsewhenc = d and A = SS* forsomeS. However,
for(A + cd*)thas notyetbeengiven.
generalexpression
for(A + cd*)twhichcoverall possible
In thispaper,wepresent
expressions
for(A + cd*)tareoftheform
eachofourexpressions
cases.Moreover,
(A + cd*)t = At + G,
ofA, At,c, d,andtheir
whereG is a matrix
obtainedfrom
onlysumsandproducts
so thatourexpressions
maybe usedto obtaintheMooreconjugate
transposes,
matrixin muchthesameway(1.1)is usedin the
Penroseinverseofa modified
case.
nonsingular
A and twocolumnsc and d,
Fora givenm x n complexmatrix
2. Notation.
notation:
weadoptthefollowing
(-)
thecomplexconjugate,
theconjugatetranspose,
-the
Moore-Penrose
inverse,
t
IL-I theEuclideannorm,
R()-the rangeorcolumnspace,
k -the columnAtc,
h -the rowd*At,
*
u
v
-the column(I -AAt)c,
-the rowd*(I - AtA),
B -the scalar1 + d*Atc.
* Receivedby the editorsMarch 6, 1972,and in revisedformJuly14, 1972.
t Departmentof Mathematics,NorthCarolina State University,
Raleigh,NorthCarolina 27607.
315
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316
CARL D. MEYER, JR.
Throughout,we make use of the factthat fora nonzero vectorx, its MoorePenroseinverseis givenby
_t =
X
=
x*
llxll2
ofthematrix(A + cd*)t,
thestructure
3. Main results.In orderto investigate
thereare six distinctcases to considerand we enumeratethemas follows:
(i) c 0 R(A) and d 0 R(A*);
(ii) c e R(A) and d R(A*) andf, = 0;
and ,B# 0;
(iii) c e R(A) and d arbitrary
(iv) c R(A) and deR(A*) and /=0;
and d e R(A*) and ,B# 0;
(v) c arbitrary
(vi) c e R(A) and d e R(A*) and /B= 0.
THEOREM 1.
(A + cd*)t = At - kut- vth+ f3vtut,
(3.1)
whenc 0 R(A) and d 0 R(A*).
THEOREM 2.
(A + cd*)t = At - kktAt- vth,
(3.2)
whenc e R(A), d 0 R(A*) and ,B= 0.
THEOREM 3. Let
P
,
1112*
ilk
Pi
V
v*-k,
=-
ql
I=
v112
k*At-h,
and
+ 1312.
Ilk11211VII2
a,=
Then,
(A + cd*)t
(3.3)
-
At + pv*k*At
a
/3~ ~
# 0.
whenc e R(A) and /3
plql,
THEOREM 4.
(A + cd*)t = At -Athth
(3.4)
whenc R(A), deR(A*) and/, = 0.
THEOREM 5. Let
P2
P2 =-
1u112
Ath* -k,
q2 -
-
kut,
11h
112
U
h
and
2=
+ 1312.
1lh
11211u112
Then,
(3.5)
(A + cd*)t
-
At + -Ath*u* - _P2q*2
whendeR(A*)and/3 0.
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317
MODIFIED MATRICES
THEOREM
6.
(A + cd*)t = At - kktAt- Athth+ (ktAtht)kh,
(3.6)
whenc E R(A), dE R(A*) and / = 0.
Beforeproving
thetheorems,
twopreliminary
factsareneeded,and we state
theseas lemmas.
LEMMA1.
[
rank(A + cd*)= rank
-1.
Proof.Thisfollows
fromthefactorization
immediately
A +
L
cd*
c
I
I
_1
][A
u$I
h 1
lv
k
-0jl I
I
0
d* 1_j
It is worthnotingthatc E R(A) ifand onlyifu = 0 and d E R(A*)ifand onlyif
v = 0.
LEMMA 2. If M and X are matricessuchthatXMMt = X and MtM = XM,
thenX = Mt.
Proof.
Mt = (MtM)Mt = XMMt = X.
We now proceedwiththeproofof thetheorems.
Throughout,
we assume
c = 0 andd # 0.
Proofof Theorem1.Let X1 denotetheright-hand
side of(3.1) and letM = A
+cd*. The proofis showingthatX1 satisfies
thefourPenroseconditions:(1)
MX1M = M, (2) XjMX1 = X1, (3) (MX1)* = MX1, and (4) (X1M)* = XjM.
UsingAvt = 0, d*vt= 1,d*k = B- 1,and c - Ak = u, it is easy to see that
MX1 = AAt + uut
so that (3) holds. Using utA = 0, utc = 1, hc = /3-1, and d* - hA = v, one
obtains
X1M = AtA + vfv
andhence(4) holds.Conditions
(1) and(2) arenoweasilyverified.
2. Let X2 denotetheright-hand
Proofof Theorem
side of (3.2). By using
Ak = c, Avt = 0, d*vt= 1,and d*k= -1 it is seen that
(A +
cd*)X2
= AAt,
whichis Hermitian.From thefactsthatktAtA= kt,hc =-1,
itfollows
that
and d* - hA = v,
X2(A + cd*) = AtA - kkt+ vtv,
whichis also Hermitian.
The first
and secondPenroseconditions
arenoweasily
verified.
Proofof Theorem
3. Thiscase is themostdifficult.
In thiscase,c E R(A)and
henceitfollowsthatR(A + cd*)c R(A)and u = 0. Since,B# 0, itis clearfrom
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318
CARL D. MEYER, JR.
Lemma 1 that
rank(A + cd*) = rank(A)
so thatR(A + cd*) = R(A) and therefore
(3.7)
(A + cd*)(A + cd*)t = AAt
because AAt is the unique orthogonalprojectoronto R(A). (For a discussionof
projectors,see [7, p. 106].) Let X3 denote the right-handside of (3.3). Because
q*AAt = ql*,it is immediatefrom(3.7) that
X3(A + cd*)(A + cd*)t = X3
and hencethefirstconditionof Lemma 2 is satisfied.
To show that the second conditionof Lemma 2 is also satisfied,we first
show that
(A + cd*)t(A + cd*) = AtA - kkt+ PiP'.
The matrixAtA - kkt+ PiPt is Hermitianand idempotent.The factthat it is
Hermitianis clear and thefactthatit is idempotentfollowsby directcomputation
usingAtAk = k,AtAp1 = - k,and kktp = - k. Since therankofan idempotent
matrixis equal to itstrace([9, p. 224]) and sincetraceis a linearfunction,
itfollows
that
rank(AtA - kkt+ PiPt) = trace(AtA - kkt+ PiPt)
-
trace(AtA) - trace(kkt)+ trace(p1pt).
Now, kktand PiPt are idempotentmatricesof rank = trace = 1 and AtA is an
idempotentmatrixwhose rankis equal to rank(A), so that
(3.8)
rank(AtA - kkt+ PtPI) = rank(A + cd*).
Using thefactsAk = c, Ap1 = -c, d*k = /-1,
d* - v,one obtains
d*pl = 1 -
l-',
and d*AtA
(A + cd*)(AtA - kkt+ PtPI) = A + cd* -c(v + fkt + cli-lpt).
-2 and hence
Now, IIp,112= llkll2'1llfl
-2, so that 1P-1Ilp -2 = f3llkll
'i-flpt
= -v
+ f,IIkIK-2p*
-
flkt.
Thus,
(A + cd*)(AtA - kkt+ PiPt) = A + cd*.
Because AtA - kkt+ PiPt is an orthogonalprojector,it followsthat
R(A* + dc*) c R(AtA - kkt+ plpt)
By virtueof(3.8), we concludethat
R(A* + dc*) = R(AtA - kkt+ plpt),
and hence,(A* + dc*)(A* + dc*)t = AtA - kkt+ PiPtI or equivalently,
(A + cd*)t(A + cd*) = AtA - kkt+ pipt.
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319
MODIFIED MATRICES
To showthatX3(A + cd*) = AtA + plpl - kkt,we computeX3(A + cd*)
afterobservingthatk*AtA
k*,q*c = 1-- a131, andq*A + d* =-11vl12f-lk*
+ v.Now,
-
X3(A + cd*) = AtA + iv*k*-
1p1q*A + (k +
= AtA + i v*k*-
p1q'*A - pid*-
= AtA + iv*k*-
-pl(ql*A+ d*)
/3
=
AtA + =v*k*-
v* d*
plqcd*
pid* + pid*
pi (v - I 'Iv2flk*).
Writev as v =-PIlkl - 2(p* + k*) and substitute
in parenthisin theexpression
I- 2 = 1fl12aL
-2 to obtain
thesesandusethefactthatlip1
1IlkIl
1
X3(A + cd*) = AtA + gSv*k*+ PiPt +
V
Since
/v* + IIkl12P1 = jv*
--
-
k112k=1
k,
k112
wearriveat
X3(A + cd*) = AtA + plpf - kkt.
Thus
(A + cd*)t(A+ cd*) = X3(A + cd*)
so thatX3 = (A + cd*)t,fromLemma2.
4. Thiscase is thedual ofTheorem2 in thesensethatit
Proofof Theorem
follows
byconsidering
conjugatetransposes
and usingthefactthatMt* = M*t
foreverymatrix
M. FromTheorem2,
(A* + dc*)t = A*t - h*h*tA*t- u*tk*
andhence,
(A + cd*)t = (A + cd*)*-* = (A*t - h*h*tA*t- u*tk*)*= At - Athth- kut.
5. Thiscase is thedual ofTheorem3 in thesensethatit
Proofof Theorem
theconjugatetranspose
followsfromTheorem3 byconsidering
ofA + cd*in a
manner
similarto thatusedintheproofofTheorem4.
Proofof Theorem6. Each ofthematrices
AAt - hth and AtA - kkt
followsfrom
is an orthogonalprojector.The factthattheyare idempotent
AAtht= ht, hAAt = h, AtAk = k and ktAtA= kt. It is clear that each is
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320
CARL D. MEYER, JR.
Hermitian.Moreover,the rank of each is equal to its traceand hence each has
rank equal to rank(A) - 1. Also, since u = 0, v = 0, and ,B= 0, it followsfrom
Lemma 1 that
rank(A + cd*) = rank(A) - 1.
Hence,
(3.9)
rank(A + cd*) = rank(AAt - hth)= rank(AAt - ktk).
WiththefactsAAtc = c, hc =-
1, and hA = d*,it is easy to see that
(AAt - hth)(A+ cd*) = (A + cd*),
so that R(A + cd*) c R(AAt - hth). Likewise,using d*AtA = d*, d*k =-1,
and Ak = c, one sees that
(A + cd*)(AtA - kkt)= A + cd*,
and henceR(A* + dc*) c R(AtA - kkt).By virtueof(3.9), it now followsthat
(3.10)
(A + cd*)(A + cd*)t
-
AAt - hth
and
(3.11)
(A + cd*)t(A+ cd*) = AtA - kkt.
If X6 denotesthe right-handside of (3.6), use (3.10) and the factthathAAt= h
to obtain
X6(A + cd*)(A + cd*)t =
X6,
whichis thefirstconditionofLemma 2. Use ktAtA= kl, hA = d*,and hc =-1
to obtain
X6(A + cd*) = AtA - kkt.
By virtueof(3.11),we have thatthesecondconditionofLemma 2 is satisfiedand
henceX6 = (A + cd*)t.
COROLLARY (The analogue of(1.1)). Whenc E R(A), d E R(A*), and ,B# 0, the
Moore-Penroseinverseof A + cd* is givenby
(A + cd*)t = At-
Atcd*At = At-
kh.
Proof. This is obtainedfromTheorem3 by settingv = 0 or fromTheorem5
by settingu = 0.
4. Special cases. In manyapplications,particularly
in statisticalapplications,
one deals not so muchwiththe Moore-Penrose inverse,but ratherwitha nonunique generalizedinverse which satisfiesonly the firstPenrose condition:
i.e., fora givenmatrixA, A- is called a generalizedinverse(g-inverse)forA if
AA-A = A. It is well known [7, p. 40] thatif 1 + d*A-c = ,B# 0 forsome Aand eitherc E R(A) or d E R(A*), thena g-inverseforA + cd* is givenby
(A + cd*)- = A
-
-'lA-cd*A-
(theanalogue of(1.1)).
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321
MODIFIED MATRICES
However,expressions
forg-inverses
of A + cd* in theothercases have never
beengiven.It is notdifficult
to examineeach oftheexpressions
for(A + cd*)t
givenearlierand decidewhichtermsof each expression
can be omittedand
whichshouldbe kept,so thatjust thefirstPenroseconditionis satisfied.
We
statetheseobservations
below.
THEOREM 7. Let A be an m x n complexmatrixand let c and d be m x 1 and
n x 1 columns,respectively.
Let A - be someg-inverse
for A; let E and F denote
matrices
E = I-AA-
and F = I-A-A;
and let/3= 1 + d*A- c. A g-inverse
forA + cd* is givenbythefollowing:
(A + cd*)-
=
A-
-
A-cc*E
Fdc*E
Fdd*A-
c*Ec
-
d*Fd + /(c*Ec)(d*Fd)'
whenc 0 R(A), dq R(A*);
(A + cd*)- = A- - Fdd*A,
(A+cd*<~A~d*Fd
when/3=0,
c e R(A), dq R(A*);
(A + cd*)- = A- --'A-cd*A-,
= O and either ceR(A)
when #0
or deR(A*);
(A + cd*)- = A - A-cc*E
c*Ec
when/3=0, c 0 R(A), d E R(A*);
(A + cd*)- = A-, when ,B= 0, c E R(A), d E R(A*).
Proof.Eachcasemaybe verified
bydirectcomputation.
5. Theorderofcomputations.
In orderto indicatehowtheresults
of? 3 may
be usedin thecomputation
oftheMoore-Penrose
inverseofa modified
matrix,
somegeneralcomments
on computations
and thefollowing
simpleexampleare
Let
provided.
A=
1
2
0
1
0
1 -1
0
O
0
1
-l-
andassumeAthasbeenpreviously
calculatedas
3
At = 112
S3
-3
I
0
-7-8A
Suppose -1 is added to the (3.3)-entryof A in order to produce the modified
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322
CARL D. MEYER, JR.
matrixA = A + cd*,where
c=
O
and d* = [O,O,-1, O].
In general,to compute t, one mustfirstdeterminewhichof the six cases
to use. Begin by computingk and h. This is always easy wheneverA arises by
a singleelementof A, because if x is added to the(i,j)th entryof A to
modifying
produceA thenk may be takenas the ithcolumnof At and h as just x timesthe
jth row ofAt. / is theneasilycomputedas
3=1+d*k
oras
3=1+hc.
In our example,,B-2/3.
In general,thenextstepis to computeu and v as
u= c
-
Ak and v = d*-
hA.
It is well known that c E R(A) if and only if u = 0, and d E R(A*) if and only if
v = 0. (Nowheredo the matrixproductsAAt or AtA need to be explicitlycomputed.)
In our example,
u=0
and v= 1[3,-1,-1,-11,
so thatTheorem3 mustbe used to compute t.
The termsappearing in Theorem 3 are now easily computedand are as
follows:
Pi
--1[-2,5,
q=
24[ 0
/
]
6-
3
v*k*At
-2_
-1
240
-2],
and
2
At= (A + cd*)t =,6
-2
2-
2
1
2
2
5
2
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MODIFIED MATRICES
323
REFERENCES
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toa changeinone
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