Products Proofs involving the Moore–Penrose inverse
1 products
1.1 has orthonormal columns
1.2 b has orthonormal rows
1.3 has full column rank , b has full row rank
1.4 conjugate transpose
products
for first 3 proofs, consider products c = ab.
a has orthonormal columns
if
a
{\displaystyle a}
has orthonormal columns i.e.
a
∗
a
=
i
{\displaystyle a^{*}a=i}
a
+
=
a
∗
{\displaystyle a^{+}=a^{*}}
. write
d
=
b
+
a
+
=
b
+
a
∗
{\displaystyle d=b^{+}a^{+}=b^{+}a^{*}}
. show
d
{\displaystyle d}
satisfies moore-penrose criteria.
c
d
c
=
a
b
b
+
a
∗
a
b
=
a
b
b
+
b
=
a
b
=
c
{\displaystyle cdc=abb^{+}a^{*}ab=abb^{+}b=ab=c}
,
d
c
d
=
b
+
a
∗
a
b
b
+
a
∗
=
b
+
b
b
+
a
∗
=
b
+
a
∗
=
d
{\displaystyle dcd=b^{+}a^{*}abb^{+}a^{*}=b^{+}bb^{+}a^{*}=b^{+}a^{*}=d}
,
(
c
d
)
∗
=
d
∗
b
∗
a
∗
=
a
(
b
+
)
∗
b
∗
a
∗
=
a
(
b
b
+
)
∗
a
∗
=
a
b
b
+
a
∗
=
c
d
{\displaystyle (cd)^{*}=d^{*}b^{*}a^{*}=a(b^{+})^{*}b^{*}a^{*}=a(bb^{+})^{*}a^{*}=abb^{+}a^{*}=cd}
,
(
d
c
)
∗
=
b
∗
a
∗
d
∗
=
b
∗
a
∗
a
(
b
+
)
∗
=
(
b
+
b
)
∗
=
b
+
b
=
b
+
a
∗
a
b
=
d
c
{\displaystyle (dc)^{*}=b^{*}a^{*}d^{*}=b^{*}a^{*}a(b^{+})^{*}=(b^{+}b)^{*}=b^{+}b=b^{+}a^{*}ab=dc}
.
therefore
d
=
c
+
{\displaystyle d=c^{+}}
.
b has orthonormal rows
if b has orthonormal rows i.e.
b
b
∗
=
i
{\displaystyle bb^{*}=i}
b
+
=
b
∗
{\displaystyle b^{+}=b^{*}}
. write
d
=
b
+
a
+
=
b
∗
a
+
{\displaystyle d=b^{+}a^{+}=b^{*}a^{+}}
. show
d
{\displaystyle d}
satisfies moore-penrose criteria.
c
d
c
=
a
b
b
∗
a
+
a
b
=
a
a
+
a
b
=
a
b
=
c
{\displaystyle cdc=abb^{*}a^{+}ab=aa^{+}ab=ab=c}
,
d
c
d
=
b
∗
a
+
a
b
b
∗
a
+
=
b
∗
a
+
a
a
+
=
b
∗
a
+
=
d
{\displaystyle dcd=b^{*}a^{+}abb^{*}a^{+}=b^{*}a^{+}aa^{+}=b^{*}a^{+}=d}
,
(
c
d
)
∗
=
d
∗
b
∗
a
∗
=
(
a
+
)
∗
b
b
∗
a
∗
=
(
a
+
)
∗
a
∗
=
(
a
a
+
)
∗
=
a
a
+
=
a
b
b
∗
a
+
=
c
d
{\displaystyle (cd)^{*}=d^{*}b^{*}a^{*}=(a^{+})^{*}bb^{*}a^{*}=(a^{+})^{*}a^{*}=(aa^{+})^{*}=aa^{+}=abb^{*}a^{+}=cd}
,
(
d
c
)
∗
=
b
∗
a
∗
d
∗
=
b
∗
a
∗
(
a
+
)
∗
b
=
b
∗
(
a
+
a
)
∗
b
=
b
∗
a
+
a
b
=
d
c
{\displaystyle (dc)^{*}=b^{*}a^{*}d^{*}=b^{*}a^{*}(a^{+})^{*}b=b^{*}(a^{+}a)^{*}b=b^{*}a^{+}ab=dc}
.
therefore
d
=
c
+
.
{\displaystyle d=c^{+}.}
a has full column rank , b has full row rank
since
a
{\displaystyle a}
has full column rank,
a
∗
a
{\displaystyle a^{*}a}
invertible
(
a
∗
a
)
+
=
(
a
∗
a
)
−
1
{\displaystyle (a^{*}a)^{+}=(a^{*}a)^{-1}}
. similarly, since
a
{\displaystyle a}
b has full row rank,
b
b
∗
{\displaystyle bb^{*}}
invertible
(
b
b
∗
)
+
=
(
b
b
∗
)
−
1
{\displaystyle (bb^{*})^{+}=(bb^{*})^{-1}}
.
write
d
=
b
+
a
+
=
b
∗
(
b
b
∗
)
−
1
(
a
∗
a
)
−
1
a
∗
{\displaystyle d=b^{+}a^{+}=b^{*}(bb^{*})^{-1}(a^{*}a)^{-1}a^{*}}
. show
d
{\displaystyle d}
satisfies moore-penrose criteria.
c
d
c
=
a
b
b
∗
(
b
b
∗
)
−
1
(
a
∗
a
)
−
1
a
∗
a
b
=
a
b
=
c
{\displaystyle cdc=abb^{*}(bb^{*})^{-1}(a^{*}a)^{-1}a^{*}ab=ab=c}
,
d
c
d
=
b
∗
(
b
b
∗
)
−
1
(
a
∗
a
)
−
1
a
∗
a
b
b
∗
(
b
b
∗
)
−
1
(
a
∗
a
)
−
1
a
∗
=
b
∗
(
b
b
∗
)
−
1
(
a
∗
a
)
−
1
a
∗
=
d
{\displaystyle dcd=b^{*}(bb^{*})^{-1}(a^{*}a)^{-1}a^{*}abb^{*}(bb^{*})^{-1}(a^{*}a)^{-1}a^{*}=b^{*}(bb^{*})^{-1}(a^{*}a)^{-1}a^{*}=d}
,
c
d
=
a
b
b
∗
(
b
b
∗
)
−
1
(
a
∗
a
)
−
1
a
∗
=
a
(
a
∗
a
)
−
1
a
∗
=
(
a
(
a
∗
a
)
−
1
a
∗
)
∗
⇒
(
c
d
)
∗
=
c
d
{\displaystyle cd=abb^{*}(bb^{*})^{-1}(a^{*}a)^{-1}a^{*}=a(a^{*}a)^{-1}a^{*}=(a(a^{*}a)^{-1}a^{*})^{*}\rightarrow (cd)^{*}=cd}
,
d
c
=
b
∗
(
b
b
∗
)
−
1
(
a
∗
a
)
−
1
a
∗
a
b
=
b
∗
(
b
b
∗
)
−
1
b
=
(
b
∗
(
b
b
∗
)
−
1
b
)
∗
⇒
(
d
c
)
∗
=
d
c
{\displaystyle dc=b^{*}(bb^{*})^{-1}(a^{*}a)^{-1}a^{*}ab=b^{*}(bb^{*})^{-1}b=(b^{*}(bb^{*})^{-1}b)^{*}\rightarrow (dc)^{*}=dc}
.
therefore
d
=
c
+
{\displaystyle d=c^{+}}
.
conjugate transpose
here,
b
=
a
∗
{\displaystyle b=a^{*}}
, ,
c
=
a
a
∗
{\displaystyle c=aa^{*}}
,
d
=
a
+
∗
a
+
{\displaystyle d=a^{+*}a^{+}}
. show indeed
d
{\displaystyle d}
satisfies 4 moore-penrose criteria.
c
d
c
=
a
a
∗
a
+
∗
a
+
a
a
∗
=
a
(
a
+
a
)
∗
a
+
a
a
∗
=
a
a
+
a
a
+
a
a
∗
=
a
a
+
a
a
∗
=
a
a
∗
=
c
{\displaystyle cdc=aa^{*}a^{+*}a^{+}aa^{*}=a(a^{+}a)^{*}a^{+}aa^{*}=aa^{+}aa^{+}aa^{*}=aa^{+}aa^{*}=aa^{*}=c}
d
c
d
=
a
+
∗
a
+
a
a
∗
a
+
∗
a
+
=
a
+
∗
a
+
a
(
a
+
a
)
∗
a
+
=
a
+
∗
a
+
a
a
+
a
a
+
=
a
+
∗
a
+
a
a
+
=
a
+
∗
a
+
a
a
+
=
a
+
∗
a
+
=
d
{\displaystyle dcd=a^{+*}a^{+}aa^{*}a^{+*}a^{+}=a^{+*}a^{+}a(a^{+}a)^{*}a^{+}=a^{+*}a^{+}aa^{+}aa^{+}=a^{+*}a^{+}aa^{+}=a^{+*}a^{+}aa^{+}=a^{+*}a^{+}=d}
(
c
d
)
∗
=
(
a
a
∗
a
+
∗
a
+
)
∗
=
a
+
∗
a
+
a
a
∗
=
a
+
∗
(
a
+
a
)
∗
a
∗
=
a
+
∗
a
∗
a
+
∗
a
∗
=
(
a
a
+
)
∗
(
a
a
+
)
∗
=
a
a
+
a
a
+
=
a
(
a
+
a
)
∗
a
+
=
{\displaystyle (cd)^{*}=(aa^{*}a^{+*}a^{+})^{*}=a^{+*}a^{+}aa^{*}=a^{+*}(a^{+}a)^{*}a^{*}=a^{+*}a^{*}a^{+*}a^{*}=(aa^{+})^{*}(aa^{+})^{*}=aa^{+}aa^{+}=a(a^{+}a)^{*}a^{+}=}
=
a
a
∗
a
+
∗
a
+
=
c
d
{\displaystyle =aa^{*}a^{+*}a^{+}=cd}
(
d
c
)
∗
=
(
a
+
∗
a
+
a
a
∗
)
∗
=
a
a
∗
a
+
∗
a
+
=
a
(
a
+
a
)
∗
a
+
=
a
a
+
a
a
+
=
(
a
a
+
)
∗
(
a
a
+
)
∗
=
a
+
∗
a
∗
a
+
∗
a
∗
=
a
+
∗
(
a
+
a
)
∗
a
∗
=
{\displaystyle (dc)^{*}=(a^{+*}a^{+}aa^{*})^{*}=aa^{*}a^{+*}a^{+}=a(a^{+}a)^{*}a^{+}=aa^{+}aa^{+}=(aa^{+})^{*}(aa^{+})^{*}=a^{+*}a^{*}a^{+*}a^{*}=a^{+*}(a^{+}a)^{*}a^{*}=}
=
a
+
∗
a
+
a
a
∗
=
d
c
{\displaystyle =a^{+*}a^{+}aa^{*}=dc}
therefore
d
=
c
+
{\displaystyle d=c^{+}}
. in other words:
(
a
a
∗
)
+
=
a
+
∗
a
+
{\displaystyle (aa^{*})^{+}=a^{+*}a^{+}}
and, since
(
a
∗
)
∗
=
a
{\displaystyle (a^{*})^{*}=a}
(
a
∗
a
)
+
=
a
+
a
+
∗
{\displaystyle (a^{*}a)^{+}=a^{+}a^{+*}}
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