Families of Commuting Normal Matrices
Denition M.1 (Notation)
i) C
n
=
_
v = (v
1
, , v
n
)
v
i
C for all 1 i n
_
ii) If C and v = (v
1
, , v
n
), w = (w
1
, , w
n
) C
n
, then
v = (v
1
, , v
n
) C
n
v +w = (v
1
+ w
1
, , v
n
+ w
n
) C
n
v, w =
n
j=1
v
j
w
j
C
The means complex conjugate.
iii) Two vectors v, w C
n
are said to be orthogonal (or perpendicular, denoted v w) if
v, w = 0.
iv) If v C
n
and A is the m n matrix whose (i, j) matrix element is A
i,j
, then Av is the
vector in C
m
with
(Av)
i
=
n
j=1
A
i,j
v
j
for all 1 i m
v) A linear subspace V of C
n
is a subset of C
n
that is closed under addition and scalar
multiplication. That is, if C and v, w V , then v, v +w V .
vi) If V is a subset of C
n
, then its orthogonal complement is
V
=
_
v C
n
v w for all w V
_
Problem M.1 Let V C
n
. Prove that V
is a linear subspace of C
n
.
Lemma M.2 Let V be a linear subspace of C
n
of dimension at least one. Let A be an nn
matrix that maps V into V . Then A has an eigenvector in V .
Proof: Let e
1
, , e
d
be a basis for V . As A maps V into itself, there exist numbers
a
i,j
, 1 i, j d such that
Ae
j
=
d
i=1
a
i,j
e
i
for all 1 j d
1
Consequently, A maps the vector w =
d
j=1
x
j
e
j
V to
Aw =
d
i,j=1
a
i,j
x
j
e
i
so that w is an eigenvector of A of eigenvalue if and only if (1) not all of the x
i
s are zero
and (2)
d
i,j=1
a
i,j
x
j
e
i
=
d
i=1
x
i
e
i
d
j=1
a
i,j
x
j
= x
i
for all 1 i d
j=1
_
a
i,j
i,j
_
x
j
= 0 for all 1 i d
For any given , the linear system of equations
d
j=1
_
a
i,j
i,j
_
x
j
= 0 for all 1 i d
has a nontrivial solution (x
1
, , x
d
) if and only if the d d matrix
_
a
i,j
i,j
1i,jd
fails to be invertible and this, in turn, is the case if and only if det
_
a
i,j
i,j
= 0. But
det
_
a
i,j
i,j
= 0 is a polynomial of degree d in and so always vanishes for at least one
value of .
Denition M.3 (Commuting) Two n n matrices A and B are said to commute if
AB = BA.
Lemma M.4 Let n 1 be an integer, V be a linear subspace of C
n
of dimension at least
one and let F be a nonempty set of n n mutually commuting matrices that map V into V .
That is, A, B F AB = BA and A F, w V Aw V . Then there exists a nonzero
vector v V that is an eigenvector for every matrix in F.
Proof: We shall show that
There is a linear subspace W of V of dimension at least one, such that each A F
is a multiple of the identity matrix when restricted to W.
This suces to prove the lemma. The proof will be by induction on the dimension d of V . If
d = 1, we may take W = V , since the restriction of any matrix to a one dimensional vector
space is a multiple of the identity.
Suppose that the claim has been proven for all dimensions strictly less than d. If
every A F is a multiple of the identity, when restricted to V , we may take W = V and we
are done. If not, pick any A F that is not a multiple of the identity when restricted to V .
2
By Lemma M.2, it has at least one eigenvector v V . Let be the corresponding eigenvalue
and set
V
= V
_
w C
n
Aw = w
_
Then V
is a linear subspace of V of dimension strictly less than d (since A, restricted to V ,
is not 1l). We claim that every B F maps V
into V
. To see this, let B F and w V
and set w
= Bw. We wish to show that w
V
. But
Aw
= ABw = BAw (A and B commute)
= Bw (Denition of V
)
= Bw = w
so w
is indeed in V
. We have veried that V
has dimension at least one and strictly smaller
than d and that every B F maps V
into V
. So we may apply the inductive hypothesis
with V replaced by V
.
Denition M.5 (Adjoint) The adjoint of the r c matrix A is the c r matrix
A
i,j
= A
j,i
Problem M.2 Let A and B be any n n matrices. Prove that B = A
if and only if
Bv, w = v, Aw for all v, w C
n
.
Problem M.3 Let A be any n n matrix. Let V be any linear subspace of C
n
and V
its
orthogonal complement. Prove that if AV V (i.e. w V Aw V ), then A
V
V
.
Denition M.6 (Normal, SelfAdjoint, Unitary)
i) An n n matrix A is normal if AA
= A
A. That is, if A commutes with its adjoint.
ii) An n n matrix A is selfadjoint if A = A
.
iii) An n n matrix U is unitary if UU
= 1l. Here 1l is the n n identity matrix. Its (i, j)
matrix element is one if i = j and zero otherwise.
Problem M.4 Let A be a normal matrix. Let be an eigenvalue of A and V the eigenspace
of A of eigenvalue . Prove that V is the eigenspace of A
of eigenvalue
.
Problem M.5 Let A be a normal matrix. Let v and w be eigenvectors of A with dierent
eigenvalues. Prove that v w.
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Problem M.6 Let A be a self-adjoint matrix. Prove that
a) A is normal
b) Every eigenvalue of A is real.
Problem M.7 Let U be a unitary matrix. Prove that
a) U is normal
b) Every eigenvalue of U obeys || = 1, i.e. is of modulus one.
Theorem M.7 Let n 1 be an integer. Let F be a nonempty set of n n mutually
commuting normal matrices. That is, A, B F AB = BA and A F AA
= A
A.
Then there exists an orthonormal basis {e
1
, , e
n
} of C
n
such that e
j
is an eigenvector of
A for every A F and 1 j n.
Proof: By Lemma M.4, with V = C
n
, there exists a nonzero vector v
1
that is an eigenvector
for every A F. Set e
1
=
v
1
v
1
and V
1
=
_
e
1
C
_
. By Problem M.4, e
1
is also an
eigenvector of A
for every A F, so A
V
1
V
1
for all A F. By Problem M.3, AV
1
V
1
for all A F.
By Lemma M.4, with V = V
1
, there exists a nonzero vector v
2
V
1
that is an
eigenvector for every A F. Choose e
2
=
v
2
v
2
. As e
2
V
1
, e
2
is orthogonal to e
1
. Dene
V
2
=
_
1
e
1
+
2
e
2
1
,
2
C
_
. By Problem M.4, e
2
is also an eigenvector of A
for every
A F, so A
V
2
V
2
for all A F. By Problem M.3, AV
2
V
2
for all A F.
By Lemma M.4, with V = V
2
, there exists a nonzero vector v
3
V
2
that is an
eigenvector for every A F. Choose e
3
=
v
3
v
3
. As e
3
V
2
, e
3
is orthogonal to both e
1
and e
2
. And so on.
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