We study various conditions on matrices $B$ and $C$ under which they can be
the off-diagonal blocks of a partitioned normal matrix.Comment: 7 page

Bhatia, Rajendra

Choi, Man-Duen

English

arXiv

We study various conditions on matrices B and C under which they can be the off-diagonal blocks of a partitioned normal matrix

Publications of the IAS Fellows

Proc. Indian Acad. Sci. (Math. Sci.) Vol. 116, No. 4, November 2006, pp. 393–399.
© Printed in India
Corners of normal matrices
RAJENDRA BHATIA and MAN-DUEN CHOI∗
Theoretical Statistics and Mathematics Unit, Indian Statistical Institute,
New Delhi 110 016, India
∗Department of Mathematics, University of Toronto, Toronto M5S 2E4, Canada
E-mail: rbh@isid.ac.in; choi@math.toronto.edu
To Kalyan Sinha on his sixtieth birthday
Abstract. We study various conditions on matrices B and C under which they can be
the off-diagonal blocks of a partitioned normal matrix.
Keywords. Normal matrix; unitary matrix; norm; completion problem; dilation.
The structure of general normal matrices is far more complicated than that of two special
kinds — hermitian and unitary. There are many interesting theorems for hermitian and
unitary matrices whose extensions to arbitrary normal matrices have proved to be extremely
recalcitrant (see e.g., [1]). The problem whose study we initiate in this note is another one
of this sort.
We consider normal matrices N of size 2n, partitioned into blocks of size n as
N =
[
A B
C D
]
. (1)
Normality imposes some restrictions on the blocks. One such restriction is the equality
‖B‖2 = ‖C‖2 (2)
between the Hilbert–Schmidt (Frobenius) norms of the off-diagonal blocks B and C. If T
is any m × m matrix with entries tij , then
‖T ‖2 =
(
m∑
j=1
|tij |2
)1/2
.
The equality (2) is a consequence of the fact that the Euclidean norm of the j th column of
a normal matrix is equal to the Euclidean norm of its j th row.
Replacing the Hilbert–Schmidt norm by another unitarily invariant norm, we may ask
whether the equality (2) is replaced by interesting inequalities. Let s1(T ) ≥ · · · ≥ sm(T )
be the singular values of T . Every unitarily invariant norm |||T ||| is a symmetric gauge
function of {sj (T )} (see chapter IV of [1] for properties of such norms). Much of our
concern in this note is with the special norms
‖T ‖2 = (tr T ∗T )1/2 =
(
m∑
j=1
s2j (T )
)1/2
393
394 Rajendra Bhatia and Man-Duen Choi
and
‖T ‖ = s1(T ) = supx∈Cm,‖x‖=1‖T x‖. (3)
The latter is the norm of T as a linear operator on the Euclidean space Cm. Clearly
‖T ‖ ≤ ‖T ‖2 ≤
√
m‖T ‖, (4)
for every m × m matrix T .
If the matrix N in (1) is hermitian, then C = B∗, and hence, |||C||| = |||B||| for all
unitarily invariant norms. If N is unitary, then AA∗ + BB∗ = A∗A + C∗C = I. Hence,
the eigenvalues λj satisfy the relations
λj (BB
∗) = λj (I − AA∗) = 1 − λj (AA∗)
= 1 − λj (A∗A) = λj (I − A∗A) = λj (C∗C).
Thus B and C have the same singular values, and again |||B||| = |||C||| for all unitarily
invariant norms.
This equality of norms does not persist when we go to arbitrary normal matrices, as we
will soon see. From (2) and (4) we get a simple inequality
||B|| ≤ √n ||C||. (5)
One may ask whether the two sides of (5) can be equal, and that is the first issue addressed
in this note.
When n = 2, it is not too difficult to construct a normal matrix N of the form (1) in
which ‖B‖ = √2‖C‖. One example of such a matrix is
N =

0 0
1 0
0 1
1 0
√
2 0
0 0
0 1
0 0
 . (6)
When n = 3, examples seem harder to come by. One that preserves some of the features
of (6) is given by the matrix
N =

0
√
2√
3
− 1 0
0 0
√
2√
3√
2√
3
+ 1 0 0
0 0 1
0 1 0
1 0 0
√
3 0 0
0 0 0
0 0 0
0 0
√
2√
3
+ 1
√
2√
3
− 1 0 0
0
√
2√
3
0

.
(7)
Corners of normal matrices 395
It can be seen that N is normal and plainly ‖B‖ = √3 while ‖C‖ = 1. When n = 4, it is
impossible to find such a matrix, and that is our first theorem.
The following elementary lemma (which can be verified by induction on the integer k)
is used repeatedly in our proof.
Lemma. Let V be an n-dimensional vector space and let V1, . . . , Vk be subspaces of V
the sum of whose dimensions is larger than (k − 1)n; i.e.,
k∑
j=1
dim Vj > (k − 1)n.
Then the intersection of these k subspaces is nonzero.
Theorem 1. There exists a normal matrix N of the form (1) with
‖B‖ = √n ‖C‖ (8)
if and only if n ≤ 3.
Proof. Note first that if equalities (2) and (8) hold simultaneously, then rank B must be
one and C must be unitary. So, after applying a unitary similarity by
[
C O
O I
]
, we may
assume that
N =
[
A B
I D
]
. (9)
The normality condition N∗N = NN∗ leads to two equations
A − D = A∗B − BD∗, (10)
2I = AA∗ − A∗A + BB∗ + B∗B + D∗D − DD∗. (11)
Since B is of rank one,
dim(ker B) = dim(ker B∗) = n − 1,
where dim X stands for the dimension of a space X. So, if n ≥ 3, then the dimensions of
ker B and ker B∗ add up to more than n. Hence their intersection is nonzero, and we may
choose a unit vector x in this intersection. For this vector, we obtain from (10)
(A − D)x = −BD∗x, (12)
and
(A − D)∗x = B∗Ax. (13)
Equation (11) leads to the condition
2 = ‖A∗x‖2 − ‖Ax‖2 + ‖Dx‖2 − ‖D∗x‖2. (14)
The rest of the proof shows that if n > 3, then we can choose a vector x ∈ (ker B) ∩
(ker B∗) for which these conditions cannot be satisfied.
396 Rajendra Bhatia and Man-Duen Choi
The two matrices BD∗ and B∗A have rank at most 1, so their kernels have dimension
at least n − 1. Hence
dim(ker B) + dim(ker B∗) + dim(ker BD∗) + dim(ker B∗A) ≥ 4n − 4.
(15)
This is larger than 3n whenever n > 4. So, in this case the four kernel spaces involved
in (15) have a nonzero intersection. Let x be a unit vector in this intersection. Then from
(12) and (13) we find that
(A − D)x = 0 and (A − D)∗x = 0.
Hence, ‖Ax‖ = ‖Dx‖ and ‖A∗x‖ = ‖D∗x‖. This contradicts the condition (14).
Now consider the case n = 4. The spaces ker B and ker B∗ have dimension 3 each,
while the space ker B(A + D)∗ has dimension at least 3. The three dimensions add up to
more than 8. Hence, we can find a unit vector x in the intersection of these three spaces.
For this vector we have
‖A∗x‖2 − ‖D∗x‖2 = Re 〈(A + D)∗x, (A − D)∗x〉
= Re 〈(A + D)∗x, B∗Ax〉
= Re 〈B(A + D)∗x,Ax〉
= 0. (16)
Here the second equality is a consequence of (13), and at the last step we have used the
fact that B(A + D)∗x = 0.
Using (12) instead of (13) we get
‖Dx‖2 − ‖Ax‖2 = Re 〈(A + D)x, (D − A)x〉
= Re 〈(A + D)x, BD∗x〉
= Re 〈B∗(A + D)x,D∗x〉. (17)
Since B is a matrix with rank equal to 1 and norm equal to 2, we have B∗BB∗ = 4B∗. (Use
the polar decomposition B = UP. In some orthonormal basis P is diagonal with only one
nonzero entry 2 on the diagonal. So B∗BB∗ = P 3U∗ = 4PU∗ = 4B∗.) Hence we have
4B∗Ax = B∗BB∗Ax
= B∗B(A − D)∗x (using (13))
= B∗B(A + D)∗x − 2B∗BD∗x
= −2B∗BD∗x
= 2B∗(A − D)x (using (12))
= 4B∗Ax − 2B∗(A + D)x.
This shows that B∗(A + D)x = 0, and we get from (17)
‖Dx‖2 − ‖Ax‖2 = 0. (18)
Corners of normal matrices 397
Clearly the relations (14), (16) and (18) cannot be simultaneously true.
We have shown that when n ≥ 4, there cannot exist a 2n × 2n normal matrix of the
form (9) in which B is an n × n matrix of rank one. This proves the theorem.
Our discussion leads to some natural questions.
Problem 1. For n ≥ 4, evaluate the quantity
αn = sup
{
‖B‖/‖C‖: ∃A,D for which
[
A B
C D
]
is normal
}
.
We have seen αn <
√
n for n ≥ 4. It would be of interest to know whether αn is a bounded
sequence.
Problem 2. What matrix pairs B,C can be the off-diagonal entries of a normal matrix N
as in (1)? In other words, when does
[
? B
C ?
]
have a normal completion?
Example 1. Consider the 2 × 2 matrices
B =
[1 ε
0 0
]
, C =
[1 0
0 ε
]
.
Then, ‖B‖2 = ‖C‖2. However, there do not exist any 2 × 2 matrices A and D for which[
A B
C D
]
is normal. We leave the verification of this statement to the reader. Thus the equality
(2) is only a necessary condition for normality of the matrix (1).
We consider some special cases of the question raised in Problem 2. We assume either
B = C, or B = C∗.
For every B, the matrix
[
? B
B ?
]
has a normal completion, and this completion may be
chosen to be of the special type
[
A B
B A
]
. Indeed, if U is the unitary matrix U = 1√
2
[
I I
−I I
]
,
then
U
[
A B
B A
]
U∗ =
[
A + B 0
0 A − B
]
.
So
[
A B
B A
]
is normal if and only if
[
A+B 0
0 A−B
]
is normal, and this is the case if and only
if A + B and A − B both are normal. The most obvious choice of A that assures this is
A = B∗. Thus
B˜ =
[
B∗ B
B B∗
]
(19)
is a normal completion of
[
? B
B ?
]
. We have the norm inequality
‖B‖ ≤ ‖B˜‖ ≤ 2‖B‖. (20)
When B =
[
0 1
0 0
]
we have ‖B˜‖ = ‖B‖. On the other hand, if B is any hermitian matrix,
then ‖B˜‖ = 2‖B‖. In this case, and more generally when B is normal,
[
0 B
B 0
]
is normal
and has norm equal to ‖B‖. This raises the question of finding completions of
[
? B
B ?
]
that
are ‘optimal’ in various senses.
398 Rajendra Bhatia and Man-Duen Choi
Problem 3. Given a matrix B find a matrix A such that
N =
[
A B
B A
]
is normal and has the least possible norm. This is equivalent to asking for a matrix A such
that A+B and A−B are normal and the quantity max(‖A+B‖, ‖A−B‖) is minimised.
It might be difficult to find all solutions to this problem. The following considerations lead
to one solution.
We assume that B is a contraction, i.e. ‖B‖ ≤ 1 and ask for an A so that
[
A B
B A
]
is
unitary. This is a unitary completion of the matrix
[
? B
B ?
]
. Let B = USV be the singular
value decomposition of B. Then[
U∗ 0
0 U∗
] [
A B
B A
] [
V ∗ 0
0 V ∗
]
=
[
U∗AV ∗ S
S U∗AV ∗
]
.
So, the problem reduces to finding an A′ such that
[
A′ S
S A′
]
is unitary. A familiar idea from
the theory of unitary dilations (p. 232 of [2]) suggests the choice A′ = i(I − S2)1/2.
This tells us how to find for any matrix B one of the least-norm normal completions of[
? B
B ?
]
. Assume ‖B‖ = 1 and find a unitary completion as proposed above.
Next we consider the case B = C∗, and ask for matrices A and D such that
N =
[
A B
B∗ D
]
(21)
is normal. A calculation shows that the matrices A and D must be normal and satisfy the
equation
(A − A∗)B = B(D − D∗). (22)
Let A = H1 + iK1 and D = H2 + iK2 be the Cartesian decompositions of A and D. Here
(H1,K1) and (H2,K2) are two pairs of commuting hermitian matrices. Equation (22) is
equivalent to K1B = BK2. This shows that
B∗BK2 = B∗K1B = (K1B)∗B = (BK2)∗B = K2B∗B.
So K2 commutes with B∗B, and hence with the factor P in the polar decomposition
B = UP.
Thus the general solution to (22) is obtained as follows: Choose K0 and K2, both
hermitian, satisfying the conditions
K0P = PK0, K2P = PK2, (K0 − K2)P = 0.
Let K1 = UK0U∗. This condition ensures
K1B = UK0U∗B = UK0P = UK2P = UPK2 = BK2.
Choose hermitian matrices H1 and H2 that commute with K1 and K2, respectively. Let
A = H1 + iK1 and D = H2 + iK2. This leads to N in (21) being normal.
Corners of normal matrices 399
As before, we also consider the special case ‖B‖ ≤ 1 and ask for A and D such that
the matrix (21) is unitary. This can be solved as follows: Let B = UP be any polar
decomposition. Choose hermitian matrices K0 and K2 that commute with P and satisfy
the inequalities
K20 ≤ I − P 2, K22 ≤ I − P 2.
Then choose hermitian matrices H0 and H2 that commute with K0 and K2, respectively,
and satisfy the conditions
H 20 + K20 = H 22 + K22 = I − P 2.
Let A = U(H0 + iK0)U∗ and D = H2 + iK2. Then the matrix (21) is unitary.
Example 1 shows that the equality ‖B‖2 = ‖C‖2 is not a sufficient condition for the
existence of a normal completion of
[
? B
C ?
]
.
Our next proposition shows that equality between all unitarily invariant norms is a
sufficient condition.
PROPOSITION
Let B,C be n× n matrices with |||B||| = |||C||| for every unitarily invariant norm. Then
the matrix
[
? B
C ?
]
has a completion that is a scalar multiple of a unitary matrix.
Proof. If |||B||| = |||C||| for every unitarily invariant norm, then sj (B) = sj (C) for all
j = 1, 2, . . . , n. Hence, there exist unitary matrices U1, U2, V1, V2 such that B = U1SU2,
and C = V1SV2. Divide B and C by ||S||, and thus assume ||S|| = 1. Then I − S2 is
positive, and has a positive square root. It is easy to see that the matrix[
(I − S2) 12 S
S −(I − S2) 12
]
is unitary. Multiply this matrix on the left by the unitary matrix U1 ⊕ V1, and on the right
by the unitary matrix V2 ⊕ U2. This gives a unitary matrix whose off-diagonal blocks are
B and C.
While the condition in the Proposition is not necessary, it is sensitive to small per-
turbations. The matrices B and C in Example 1 satisfy the conditions ‖B‖2 = ‖C‖2,
|||B||| = |||C||| +O(ε), but for ε = 0, there is no possible normal completion of
[
? B
C ?
]
.
Acknowledgement
The second author thanks the Indian Statistical Institute and NSERC of Canada for sup-
porting a visit to New Delhi during which this work was initiated.
References
[1] Bhatia R, Matrix Analysis (Springer) (1997)
[2] Halmos P R, A Hilbert Space Problem Book, 2nd edition (Springer) (1982)


Corners of normal matrices

Indian Academy of Sciences

Proc. Indian Acad. Sci. (Math. Sci.) Vol. 116, No. 4, November 2006, pp. 393–399.
© Printed in India
Corners of normal matrices
RAJENDRA BHATIA and MAN-DUEN CHOI∗
Theoretical Statistics and Mathematics Unit, Indian Statistical Institute,
New Delhi 110 016, India
∗Department of Mathematics, University of Toronto, Toronto M5S 2E4, Canada
E-mail: rbh@isid.ac.in; choi@math.toronto.edu
To Kalyan Sinha on his sixtieth birthday
Abstract. We study various conditions on matrices B and C under which they can be
the off-diagonal blocks of a partitioned normal matrix.
Keywords. Normal matrix; unitary matrix; norm; completion problem; dilation.
The structure of general normal matrices is far more complicated than that of two special
kinds — hermitian and unitary. There are many interesting theorems for hermitian and
unitary matrices whose extensions to arbitrary normal matrices have proved to be extremely
recalcitrant (see e.g., [1]). The problem whose study we initiate in this note is another one
of this sort.
We consider normal matrices N of size 2n, partitioned into blocks of size n as
N =
[
A B
C D
]
. (1)
Normality imposes some restrictions on the blocks. One such restriction is the equality
‖B‖2 = ‖C‖2 (2)
between the Hilbert–Schmidt (Frobenius) norms of the off-diagonal blocks B and C. If T
is any m × m matrix with entries tij , then
‖T ‖2 =
(
m∑
j=1
|tij |2
)1/2
.
The equality (2) is a consequence of the fact that the Euclidean norm of the j th column of
a normal matrix is equal to the Euclidean norm of its j th row.
Replacing the Hilbert–Schmidt norm by another unitarily invariant norm, we may ask
whether the equality (2) is replaced by interesting inequalities. Let s1(T ) ≥ · · · ≥ sm(T )
be the singular values of T . Every unitarily invariant norm |||T ||| is a symmetric gauge
function of {sj (T )} (see chapter IV of [1] for properties of such norms). Much of our
concern in this note is with the special norms
‖T ‖2 = (tr T ∗T )1/2 =
(
m∑
j=1
s2j (T )
)1/2
393
394 Rajendra Bhatia and Man-Duen Choi
and
‖T ‖ = s1(T ) = supx∈Cm,‖x‖=1‖T x‖. (3)
The latter is the norm of T as a linear operator on the Euclidean space Cm. Clearly
‖T ‖ ≤ ‖T ‖2 ≤
√
m‖T ‖, (4)
for every m × m matrix T .
If the matrix N in (1) is hermitian, then C = B∗, and hence, |||C||| = |||B||| for all
unitarily invariant norms. If N is unitary, then AA∗ + BB∗ = A∗A + C∗C = I. Hence,
the eigenvalues λj satisfy the relations
λj (BB
∗) = λj (I − AA∗) = 1 − λj (AA∗)
= 1 − λj (A∗A) = λj (I − A∗A) = λj (C∗C).
Thus B and C have the same singular values, and again |||B||| = |||C||| for all unitarily
invariant norms.
This equality of norms does not persist when we go to arbitrary normal matrices, as we
will soon see. From (2) and (4) we get a simple inequality
||B|| ≤ √n ||C||. (5)
One may ask whether the two sides of (5) can be equal, and that is the first issue addressed
in this note.
When n = 2, it is not too difficult to construct a normal matrix N of the form (1) in
which ‖B‖ = √2‖C‖. One example of such a matrix is
N =

0 0
1 0
0 1
1 0
√
2 0
0 0
0 1
0 0
 . (6)
When n = 3, examples seem harder to come by. One that preserves some of the features
of (6) is given by the matrix
N =

0
√
2√
3
− 1 0
0 0
√
2√
3√
2√
3
+ 1 0 0
0 0 1
0 1 0
1 0 0
√
3 0 0
0 0 0
0 0 0
0 0
√
2√
3
+ 1
√
2√
3
− 1 0 0
0
√
2√
3
0

.
(7)
Corners of normal matrices 395
It can be seen that N is normal and plainly ‖B‖ = √3 while ‖C‖ = 1. When n = 4, it is
impossible to find such a matrix, and that is our first theorem.
The following elementary lemma (which can be verified by induction on the integer k)
is used repeatedly in our proof.
Lemma. Let V be an n-dimensional vector space and let V1, . . . , Vk be subspaces of V
the sum of whose dimensions is larger than (k − 1)n; i.e.,
k∑
j=1
dim Vj > (k − 1)n.
Then the intersection of these k subspaces is nonzero.
Theorem 1. There exists a normal matrix N of the form (1) with
‖B‖ = √n ‖C‖ (8)
if and only if n ≤ 3.
Proof. Note first that if equalities (2) and (8) hold simultaneously, then rank B must be
one and C must be unitary. So, after applying a unitary similarity by
[
C O
O I
]
, we may
assume that
N =
[
A B
I D
]
. (9)
The normality condition N∗N = NN∗ leads to two equations
A − D = A∗B − BD∗, (10)
2I = AA∗ − A∗A + BB∗ + B∗B + D∗D − DD∗. (11)
Since B is of rank one,
dim(ker B) = dim(ker B∗) = n − 1,
where dim X stands for the dimension of a space X. So, if n ≥ 3, then the dimensions of
ker B and ker B∗ add up to more than n. Hence their intersection is nonzero, and we may
choose a unit vector x in this intersection. For this vector, we obtain from (10)
(A − D)x = −BD∗x, (12)
and
(A − D)∗x = B∗Ax. (13)
Equation (11) leads to the condition
2 = ‖A∗x‖2 − ‖Ax‖2 + ‖Dx‖2 − ‖D∗x‖2. (14)
The rest of the proof shows that if n > 3, then we can choose a vector x ∈ (ker B) ∩
(ker B∗) for which these conditions cannot be satisfied.
396 Rajendra Bhatia and Man-Duen Choi
The two matrices BD∗ and B∗A have rank at most 1, so their kernels have dimension
at least n − 1. Hence
dim(ker B) + dim(ker B∗) + dim(ker BD∗) + dim(ker B∗A) ≥ 4n − 4.
(15)
This is larger than 3n whenever n > 4. So, in this case the four kernel spaces involved
in (15) have a nonzero intersection. Let x be a unit vector in this intersection. Then from
(12) and (13) we find that
(A − D)x = 0 and (A − D)∗x = 0.
Hence, ‖Ax‖ = ‖Dx‖ and ‖A∗x‖ = ‖D∗x‖. This contradicts the condition (14).
Now consider the case n = 4. The spaces ker B and ker B∗ have dimension 3 each,
while the space ker B(A + D)∗ has dimension at least 3. The three dimensions add up to
more than 8. Hence, we can find a unit vector x in the intersection of these three spaces.
For this vector we have
‖A∗x‖2 − ‖D∗x‖2 = Re 〈(A + D)∗x, (A − D)∗x〉
= Re 〈(A + D)∗x, B∗Ax〉
= Re 〈B(A + D)∗x, Ax〉
= 0. (16)
Here the second equality is a consequence of (13), and at the last step we have used the
fact that B(A + D)∗x = 0.
Using (12) instead of (13) we get
‖Dx‖2 − ‖Ax‖2 = Re 〈(A + D)x, (D − A)x〉
= Re 〈(A + D)x, BD∗x〉
= Re 〈B∗(A + D)x, D∗x〉. (17)
Since B is a matrix with rank equal to 1 and norm equal to 2, we have B∗BB∗ = 4B∗. (Use
the polar decomposition B = UP. In some orthonormal basis P is diagonal with only one
nonzero entry 2 on the diagonal. So B∗BB∗ = P 3U∗ = 4PU∗ = 4B∗.) Hence we have
4B∗Ax = B∗BB∗Ax
= B∗B(A − D)∗x (using (13))
= B∗B(A + D)∗x − 2B∗BD∗x
= −2B∗BD∗x
= 2B∗(A − D)x (using (12))
= 4B∗Ax − 2B∗(A + D)x.
This shows that B∗(A + D)x = 0, and we get from (17)
‖Dx‖2 − ‖Ax‖2 = 0. (18)
Corners of normal matrices 397
Clearly the relations (14), (16) and (18) cannot be simultaneously true.
We have shown that when n ≥ 4, there cannot exist a 2n × 2n normal matrix of the
form (9) in which B is an n × n matrix of rank one. This proves the theorem.
Our discussion leads to some natural questions.
Problem 1. For n ≥ 4, evaluate the quantity
αn = sup
{
‖B‖/‖C‖: ∃A, D for which
[
A B
C D
]
is normal
}
.
We have seen αn <
√
n for n ≥ 4. It would be of interest to know whether αn is a bounded
sequence.
Problem 2. What matrix pairs B, C can be the off-diagonal entries of a normal matrix N
as in (1)? In other words, when does
[
? B
C ?
]
have a normal completion?
Example 1. Consider the 2 × 2 matrices
B =
[
1 ε
0 0
]
, C =
[
1 0
0 ε
]
.
Then, ‖B‖2 = ‖C‖2. However, there do not exist any 2 × 2 matrices A and D for which[
A B
C D
]
is normal. We leave the verification of this statement to the reader. Thus the equality
(2) is only a necessary condition for normality of the matrix (1).
We consider some special cases of the question raised in Problem 2. We assume either
B = C, or B = C∗.
For every B, the matrix
[
? B
B ?
]
has a normal completion, and this completion may be
chosen to be of the special type
[
A B
B A
]
. Indeed, if U is the unitary matrix U = 1√
2
[
I I
−I I
]
,
then
U
[
A B
B A
]
U∗ =
[
A + B 0
0 A − B
]
.
So
[
A B
B A
]
is normal if and only if
[
A+B 0
0 A−B
]
is normal, and this is the case if and only
if A + B and A − B both are normal. The most obvious choice of A that assures this is
A = B∗. Thus
B̃ =
[
B∗ B
B B∗
]
(19)
is a normal completion of
[
? B
B ?
]
. We have the norm inequality
‖B‖ ≤ ‖B̃‖ ≤ 2‖B‖. (20)
When B =
[
0 1
0 0
]
we have ‖B̃‖ = ‖B‖. On the other hand, if B is any hermitian matrix,
then ‖B̃‖ = 2‖B‖. In this case, and more generally when B is normal,
[
0 B
B 0
]
is normal
and has norm equal to ‖B‖. This raises the question of finding completions of
[
? B
B ?
]
that
are ‘optimal’ in various senses.
398 Rajendra Bhatia and Man-Duen Choi
Problem 3. Given a matrix B find a matrix A such that
N =
[
A B
B A
]
is normal and has the least possible norm. This is equivalent to asking for a matrix A such
that A+B and A−B are normal and the quantity max(‖A+B‖, ‖A−B‖) is minimised.
It might be difficult to find all solutions to this problem. The following considerations lead
to one solution.
We assume that B is a contraction, i.e. ‖B‖ ≤ 1 and ask for an A so that
[
A B
B A
]
is
unitary. This is a unitary completion of the matrix
[
? B
B ?
]
. Let B = USV be the singular
value decomposition of B. Then[
U∗ 0
0 U∗
] [
A B
B A
] [
V ∗ 0
0 V ∗
]
=
[
U∗AV ∗ S
S U∗AV ∗
]
.
So, the problem reduces to finding an A′ such that
[
A′ S
S A′
]
is unitary. A familiar idea from
the theory of unitary dilations (p. 232 of [2]) suggests the choice A′ = i(I − S2)1/2.
This tells us how to find for any matrix B one of the least-norm normal completions of[
? B
B ?
]
. Assume ‖B‖ = 1 and find a unitary completion as proposed above.
Next we consider the case B = C∗, and ask for matrices A and D such that
N =
[
A B
B∗ D
]
(21)
is normal. A calculation shows that the matrices A and D must be normal and satisfy the
equation
(A − A∗)B = B(D − D∗). (22)
Let A = H1 + iK1 and D = H2 + iK2 be the Cartesian decompositions of A and D. Here
(H1, K1) and (H2, K2) are two pairs of commuting hermitian matrices. Equation (22) is
equivalent to K1B = BK2. This shows that
B∗BK2 = B∗K1B = (K1B)∗B = (BK2)∗B = K2B∗B.
So K2 commutes with B∗B, and hence with the factor P in the polar decomposition
B = UP.
Thus the general solution to (22) is obtained as follows: Choose K0 and K2, both
hermitian, satisfying the conditions
K0P = PK0, K2P = PK2, (K0 − K2)P = 0.
Let K1 = UK0U∗. This condition ensures
K1B = UK0U∗B = UK0P = UK2P = UPK2 = BK2.
Choose hermitian matrices H1 and H2 that commute with K1 and K2, respectively. Let
A = H1 + iK1 and D = H2 + iK2. This leads to N in (21) being normal.
Corners of normal matrices 399
As before, we also consider the special case ‖B‖ ≤ 1 and ask for A and D such that
the matrix (21) is unitary. This can be solved as follows: Let B = UP be any polar
decomposition. Choose hermitian matrices K0 and K2 that commute with P and satisfy
the inequalities
K20 ≤ I − P 2, K22 ≤ I − P 2.
Then choose hermitian matrices H0 and H2 that commute with K0 and K2, respectively,
and satisfy the conditions
H 20 + K20 = H 22 + K22 = I − P 2.
Let A = U(H0 + iK0)U∗ and D = H2 + iK2. Then the matrix (21) is unitary.
Example 1 shows that the equality ‖B‖2 = ‖C‖2 is not a sufficient condition for the
existence of a normal completion of
[
? B
C ?
]
.
Our next proposition shows that equality between all unitarily invariant norms is a
sufficient condition.
PROPOSITION
Let B, C be n × n matrices with |||B||| = |||C||| for every unitarily invariant norm. Then
the matrix
[
? B
C ?
]
has a completion that is a scalar multiple of a unitary matrix.
Proof. If |||B||| = |||C||| for every unitarily invariant norm, then sj (B) = sj (C) for all
j = 1, 2, . . . , n. Hence, there exist unitary matrices U1, U2, V1, V2 such that B = U1SU2,
and C = V1SV2. Divide B and C by ||S||, and thus assume ||S|| = 1. Then I − S2 is
positive, and has a positive square root. It is easy to see that the matrix[
(I − S2) 12 S
S −(I − S2) 12
]
is unitary. Multiply this matrix on the left by the unitary matrix U1 ⊕ V1, and on the right
by the unitary matrix V2 ⊕ U2. This gives a unitary matrix whose off-diagonal blocks are
B and C.
While the condition in the Proposition is not necessary, it is sensitive to small per-
turbations. The matrices B and C in Example 1 satisfy the conditions ‖B‖2 = ‖C‖2,
|||B||| = |||C||| + O(ε), but for ε = 0, there is no possible normal completion of
[
? B
C ?
]
.
Acknowledgement
The second author thanks the Indian Statistical Institute and NSERC of Canada for sup-
porting a visit to New Delhi during which this work was initiated.
References
[1] Bhatia R, Matrix Analysis (Springer) (1997)
[2] Halmos P R, A Hilbert Space Problem Book, 2nd edition (Springer) (1982)


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Corners of normal matrices

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