In mathematics , the Hadamard inequality describes an estimate for the determinant of a square matrix . It is named after the French mathematician Jacques Salomon Hadamard .
Classic Hadamard inequality
Let be a matrix over the complex numbers with the column vectors , then with the Euclidean norm
M.
{\ displaystyle M}
(
n
×
n
)
{\ displaystyle (n \ times n)}
m
1
,
...
,
m
n
{\ displaystyle m_ {1}, \ dots, m_ {n}}
‖
⋅
‖
2
{\ displaystyle \ | \ cdot \ | _ {2}}
|
det
M.
|
≤
∏
i
=
1
n
‖
m
i
‖
2
.
{\ displaystyle | \ det M | \, \ leq \, \ prod _ {i = 1} ^ {n} \ | m_ {i} \ | _ {2}.}
With the QR decomposition of the matrix, the following applies
M.
=
Q
R.
{\ displaystyle M = QR}
M.
{\ displaystyle M}
|
det
M.
|
=
|
det
Q
|
⋅
|
det
R.
|
=
|
det
R.
|
≤
‖
r
1
‖
2
⋅
...
⋅
‖
r
n
‖
2
,
{\ displaystyle | \ det M | = | \ det Q | \ cdot | \ det R | = | \ det R | \ leq \ | r_ {1} \ | _ {2} \ cdot \ ldots \ cdot \ | r_ {n} \ | _ {2},}
where is.
‖
r
i
‖
2
=
‖
Q
r
i
‖
2
=
‖
m
i
‖
2
{\ displaystyle \ | r_ {i} \ | _ {2} = \ | Qr_ {i} \ | _ {2} = \ | m_ {i} \ | _ {2}}
Geometric view
Is a matrix with real entries, then the volume of their row or column vectors spanned dimensional parallelepiped . This volume is maximal for orthogonal rows (or columns) and is consequently at most as large as the volume of the -dimensional cuboid with edges of the lengths .
M.
{\ displaystyle M}
(
n
×
n
)
{\ displaystyle (n \ times n)}
|
det
(
M.
)
|
{\ displaystyle | \ det (M) |}
m
i
{\ displaystyle m_ {i}}
n
{\ displaystyle n}
∏
i
=
1
n
‖
m
i
‖
2
{\ displaystyle \ prod _ {i = 1} ^ {n} \ | m_ {i} \ | _ {2}}
n
{\ displaystyle n}
‖
m
i
‖
2
{\ displaystyle \ | m_ {i} \ | _ {2}}
Attenuated Hadamard Inequality
Let be a commutative ring with pseudo magnitude and a matrix over with the row vectors . Then applies
(
R.
,
|
⋅
|
)
{\ displaystyle (R, | \ cdot |)}
M.
{\ displaystyle M}
(
n
×
n
)
{\ displaystyle (n \ times n)}
R.
{\ displaystyle R}
m
1
,
...
,
m
n
{\ displaystyle m_ {1}, \ dots, m_ {n}}
|
det
M.
|
≤
∏
i
=
1
n
‖
m
i
‖
1
{\ displaystyle | \ det M | \, \ leq \, \ prod _ {i = 1} ^ {n} \ | m_ {i} \ | _ {1}}
with the 1- pseudo norm .
Remarks
Because of this, the classical Hadamard inequality provides the sharper estimate.
‖
x
‖
2
≤
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x
‖
1
{\ displaystyle \ | x \ | _ {2} \ leq \ | x \ | _ {1}}
If a ring is based on the usual absolute value function of complex numbers (example: the whole numbers ), then the more stringent classical Hadamard inequality is always applicable.
R.
⊆
C.
{\ displaystyle R \ subseteq \ mathbb {C}}
Z
{\ displaystyle \ mathbb {Z}}
literature
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