proof of Hadamard’s inequality

proof of Hadamard’s inequality

Let’s first prove the second

inequality

. If

A

is

singular

, the thesis is
trivially verified, since for a

Hermitian

positive semidefinite matrix

the
right-hand side is always nonnegative, all the

diagonal

entries being nonnegative. Let’s thus assume

det

⁡

(

A

)

≠

0

, which means,

A

being
Hermitian

positive semidefinite

,

det

⁡

(

A

)

>

0

. Then no diagonal entry of

A

can be

0

(otherwise, since for a Hermitian positive semidefinite matrix,

0

≤

λ

m

⁢

i

⁢

n

≤

a

i

⁢

i

≤

λ

m

⁢

a

⁢

x

,

λ

m

⁢

i

⁢

n

and

λ

m

⁢

a

⁢

x

being respectively the minimal and the maximal

eigenvalue

, this would imply

λ

m

⁢

i

⁢

n

=

0

, that is

A

is singular); for this reason we can define

D

=

d

⁢

i

⁢

a

⁢

g

⁢

(

d

11

,

d

22

,

…

,

d

n

⁢

n

)

, with

d

i

⁢

i

=

a

i

⁢

i

-

1

2

∈

ℝ

, since all

a

i

⁢

i

∈

ℝ

+

. Let’s furthermore define

B

=

D

⁢

A

⁢

D

. It’s easy to check that

B

too is Hermitian positive semidefinite, so its eigenvalues

λ

B

are all non-negative (actually, since

A

=

A

H

and since

D

is real and diagonal,

B

H

=

(

D

⁢

A

⁢

D

)

H

=

D

H

⁢

(

D

⁢

A

)

H

=

D

H

⁢

A

H

⁢

D

H

=

D

⁢

A

⁢

D

=

B

; on the other hand, for any

𝐱

≠

𝟎

,

𝐱

H

⁢

B

⁢

𝐱

=

𝐱

H

⁢

D

⁢

A

⁢

D

⁢

𝐱

=

(

𝐱

H

⁢

D

)

⁢

A

⁢

(

D

⁢

𝐱

)

=

(

D

H

⁢

𝐱

)

H

⁢

A

⁢

(

D

⁢

𝐱

)

=

(

D

⁢

𝐱

)

H

⁢

A

⁢

(

D

⁢

𝐱

)

=

𝐲

H

⁢

A

⁢

𝐲

≥

0

).
Moreover, we have obviously

b

i

⁢

i

=

d

i

⁢

i

⁢

a

i

⁢

i

⁢

d

i

⁢

i

=

1

so that

t

⁢

r

⁢

(

B

)

=

n

and, recalling the geometric-arithmetic mean inequality (

http://planetmath.org/ArithmeticGeometricMeansInequality

), which holds in this
case because the eigenvalues of

B

are all non-negative,

det

⁡

(

B

)

=

∏

i

=

1

n

λ

B

(

i

)

≤

(

1

n

⁢

∑

i

=

1

n

λ

B

(

i

)

)

n

=

(

1

n

⁢

t

⁢

r

⁢

(

B

)

)

n

=

1

,

and since

det

⁡

(

B

)

=

det

⁡

(

D

)

2

⁢

det

⁡

(

A

)

=

(

∏

i

=

1

n

a

i

⁢

i

)

-

1

⁢

det

⁡

(

A

)

, we have the thesis.
Since

det

⁡

(

A

)

=

∏

i

=

1

n

a

i

⁢

i

if and only if

det

⁡

(

B

)

=

1

and since in the geometric-arithmetic inequality equality holds if
and only if all terms are equal, we must have

λ

B

(

i

)

=

λ

B

, so that

∏

i

=

1

n

λ

B

(

i

)

=

λ

B

n

=

1

,
whence

λ

B

=

1

(

λ

B

having to be non-negative), and since

B

is Hermitian and hence is

diagonalizable

, we obtain

B

=

I

, and so

A

=

D

-

1

⁢

B

⁢

D

-

1

=

D

-

2

=

d

⁢

i

⁢

a

⁢

g

⁢

(

a

11

,

a

22

,

…

,

a

n

⁢

n

)

. So we can
conclude that equality holds if and only if

A

is diagonal.

Let’s now derive the more general first inequality.
Let

A

be a complex-valued

n

×

n

matrix. If

A

is singular,
the thesis is trivially verified. Let’s thus assume

det

⁡

(

A

)

≠

0

; then

B

=

A

⁢

A

H

is a Hermitian positive semidefinite matrix (actually,

B

H

=

(

A

⁢

A

H

)

H

=

(

A

H

)

H

⁢

A

H

=

A

⁢

A

H

=

B

and, for any

𝐱

≠

𝟎

,

𝐱

H

⁢

B

⁢

𝐱

=

𝐱

H

⁢

A

⁢

A

H

⁢

𝐱

=

(

𝐱

H

⁢

A

)

⁢

(

A

H

⁢

𝐱

)

=

(

A

H

⁢

𝐱

)

H

⁢

(

A

H

⁢

𝐱

)

=

𝐲

H

⁢

𝐲

=

∥

𝐲

∥

2

2

≥

0

). Therefore, the second inequality can be applied to

B

, yielding:

|

det

⁡

(

A

)

|

2

=

det

⁡

(

A

)

⁢

det

∗

⁡

(

A

)

=

det

⁡

(

A

)

⁢

det

⁡

(

A

H

)

=

det

⁡

(

A

⁢

A

H

)

=

=

det

⁡

(

B

)

≤

∏

i

=

1

n

b

i

⁢

i

=

∏

i

=

1

n

∑

j

=

1

n

a

i

⁢

j

⁢

(

a

H

)

j

⁢

i

=

∏

i

=

1

n

∑

j

=

1

n

a

i

⁢

j

⁢

a

i

⁢

j

∗

=

∏

i

=

1

n

∑

j

=

1

n

|

a

i

⁢

j

|

2

.

As we proved above, for

det

⁡

(

B

)

to be equal to

∏

i

=

1

n

b

i

⁢

i

,

B

must be diagonal, which means that

∑

k

=

1

n

a

i

⁢

k

⁢

a

j

⁢

k

∗

=

|

a

i

⁢

i

|

2

⁢

δ

i

⁢

j

. So we can conclude that equality holds if and only if the rows of

A

are

orthogonal

.

□

References

1

R. A. Horn, C. R. Johnson,

Matrix Analysis

, Cambridge University Press, 1985

Title

proof of

Hadamard

’s inequality

Canonical name

ProofOfHadamardsInequality

Date of creation

2013-03-22 15:37:02

Last modified on

2013-03-22 15:37:02

Owner

Andrea Ambrosio (7332)

Last modified by

Andrea Ambrosio (7332)

Numerical id

17

Author

Andrea Ambrosio (7332)

Entry type

Proof

Classification

msc 15A45

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