A conclusion on determinant of positive definite real symmetric matrix
(Department of Mathematics, Yangtze Normal University, Chongqing 408 100)
Yang Shixian
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Based on the knowledge of measurement matrix and block matrix, this paper draws the following conclusions.
The difference between the determinant and its main diagonal elements of the stereotyped real symmetric matrix is obtained
Equation.
Keywords: Hermite orthogonalization block of real symmetric matrix metric matrix
Matrix determinant
Real symmetric matrix is an important content in higher algebra, so-called fixed real.
Symmetric matrices are positive definite, negative definite, semi-positive definite and semi-negative definite matrices. Let's go back first.
Consider some conclusions about real symmetric matrices that will be used in this paper:
Property 1: The necessary and sufficient condition for the positive definite of real symmetric matrix A is the existence of invertible square matrix.
The matrix c makes a = c 'c.
Property 2: Real symmetric matrix A is semi-positive definite if and only if it is all.
Both the main expression and the subordinate expression are greater than or equal to zero.
Property 3: The necessary and sufficient condition for the real symmetric matrix A to be negative definite (semi-negative definite) is-a..
Is positive definite (semi-positive definite).
Property 4: In N-dimensional Euclidean space, a set of bases ε 1, ε2,? ，εn
Metric matrix of A=
(aiji), where aiji = (ε i, ε j) is a real symmetric matrix and the matrix A is positive definite.
Property 5: In N-dimensional Euclidean space, two sets of bases ε 1, ε2,? ，εn
And η 1, η2,? ，ηn
The metric matrices are a and b, respectively, then a and b are contractual, that is, if (η 1, η2,
，ηn ) =(ε 1，ε2，？ , εn)C, then there is B=C'AC.
The main theorem to be proved in this paper is:
Theorem 1: A = (aiji) is a positive definite matrix of order n, then there is detA≤
n
k= 1
* akk
To prove the theorem 1, we first prove a lemma:
Lemma: ε 1, ε2,? ，εn
Is a set of bases of n-dimensional Euclidean space, ε 1, ε2,? ，εn
famous book
The super-ermite is orthogonalized into η 1, η2,? , ηn, remember G(ε 1, ε2,? , εn) is ε 1, ε2,
，εn
The metric matrix of, prove:
detG(ε 1
,ε2, ? ，εn)=detG(η 1，η2，？ ，ηn)=|η 1|2
|η2|2
? |ηn|2
Proof: Suppose A comes from ε 1.
,ε2
, ? ，εn
To η 1, η2,? ，ηn
Transfer matrix, namely:
(η 1 ,η2, ? ，ηn)=(ε 1，ε2，？ ，εn)A
So G(η 1, η2,? ，ηn)= A′G(ε 1，ε2，？ ，εn)A ( 1)
According to the meaning of the question η 1, η2,? ，ηn
Is composed of ε 1, ε2,? ，εn
Through Hermite orthogonalization
Come, so there are:
η 1
=ε 1;
η2
=ε2-
(ε2 ,η 1)
(η 1,η 1)
η 1
ηn
=εn-
(εn，η 1)
(η 1,η 1)
η 1
- ? -
(εn，ηn- 1)
(ηn- 1，ηn- 1)
ηn- 1
So we know that A is an upper triangular matrix, and all the elements on the main diagonal are 1, that is,
A=
1 * ? *
0 1 ? *
# ?
0 0 ?
$ % % & amp
'(()
1
At the same time, a' =
1 0 ? 0
* 1 ? 0
# ?
* * ?
$ % % & amp
'(()
1
, so detA'=
DetA= 1. According to the formula (1):
detG(η 1，η2，？ ，ηn)= det(A′G(ε 1，ε2，？ ，εn)A)
= detA′detG(ε 1，ε，？ ，εn) detA
=detG(ε 1，ε2，？ ，εn)
Because η 1, η2, ηn
Is an orthogonal vector group, so G(η 1, η2,? , ηn) is right.
Angle matrix, and:
detG(η 1，η2，？ ，ηn)=|η 1|2
|η2|2
? |ηn|2
Namely: detG(ε 1, ε2,? ，εn)=detG(η 1，η2，？ ，ηn)=|η 1| 2
|η2 | 2
| η n | 2, the certificate is over.
Proof of Theorem 1: According to the meaning of the question, A = (aiji) is a positive definite matrix of order n, so it depends on properties.
Quality 1 knows the existence of invertible square matrix c, so a = c' C. Let the N-column vectors of matrix c be
Is α 1, α2,? , αn, the fast matrices using multiplication and division are:
a = C′C =
α 1
′
α2
′
αn
$ % % % & amp
'((()
′
(α 1 ,α2, ? ，αn)=
α 1
′α 1
α 1
′α2 ? α 1′αn
α2
′α 1
α2
′α2 ? α2′αn
αn
′α 1
αn′α2
αn′αn
$ % % % & amp
'((()
=
(α 1,α 1) (α 1,α2) ? (α 1，αn)
(α2,α 1) (α2,α2) ? (α2，αn)
(αn，α 1) (αn，α2)？ (αn，αn
$ % % % & amp
'((()
)
( 2)
Because the matrix is a reversible square matrix, α 1, α2,? ，αn
Is a linearly independent direction
The quantity group can also be regarded as Rn.
A set of bases, then the matrix A is α 1, α2,? ，αn
The metric matrix of. Suppose α 1, α2,? ，αn
Hermite orthogonalization is performed to obtain vectors.
β 1，β2，？ βn is known from the condition of lemma:
det A=|β 1|2
|β2|2
? βn|2
Because β 1, β2,? ，βn
Is composed of α 1, α2,? ，αn
Through Hermite orthogonalization
Come, they have the following relationship:
β 1
=α 1 ;
β2
=α2-
(α2,β 1)
(β 1,β 1)
β 1
βn
=αn-
(αn，β 1)
(β 1,β 1)
β 1
- ? -
(αn，βn- 1)
(βn- 1，βn- 1)
βn
- 1。
Use β 1, β2,? ，βn
Represents α 1, α2,? ，αn
There are:
α 1
=β 1 ;
α2
=β2+
(α2,β 1)
(β 1,β 1)
β 1
αn
=βn+
(αn，β 1)
(β 1,β 1)
β 1
+? +
(αn，βn- 1)
(βn- 1，βn- 1)
βn
- 1
Because β 1, β2,? ，βn
Pairwise orthogonal, so there are:
|α 1|=|β 1|;
|α2|= β2+
(α2,β 1)
(β 1,β 1)
β 1
=|β2|+
(α2,β 1)
(β 1,β 1)
β 1
≥|β2|;
|αn|= βn+
(αn，β 1)
(β 1,β 1)
β 1
+? +
(αn，βn- 1)
(βn- 1，βn- 1)
βn
- 1
=|βn|+
(αn，β 1)
(β 1 ,β 1)
β 1
+? +
(αn，βn- 1)
(βn- 1，βn- 1)
βn- 1
≥|βn
|
So: det A=|β 1|2.
|β2|2
? βn|2
≤|α 1|2
≤|α 1|2
|α2|2
? |αn|2，
It is easy to know from formula (2) that |ak|2=akk,
That is, det A≤
n
k= 1
∏akk, the certificate is over.
We know that the determinant A=( aij) of a semi-positive definite matrix must be greater than or equal to.
At zero, when det A >, 0, a must be positive definite; Principal Pair of Simultaneous Semi-positive Definite Matrix A
The elements akk( 1≤k≤n) on the corner line are all non-negative real numbers, so when det A=0,
Inequality det A≤
n
k= 1
* akk
Obviously established. To sum up, the theorem 1 has:
Inference1:a = (aiji) is an n-order semi-positive definite matrix, then det A≤
n
k= 1
* akk .
For semi-positive definite or negative definite matrix A=( aij), we know that -A is semi-positive definite or.
Positive definite, so:
Inference 2: A = (aiji) is a semi-negative definite (negative definite) matrix of order n, and when n is even, there is
det A≤
n
k= 1
* akk； When n is odd, there exists det A≥
n
k= 1
* akk .
It is proved that if A =(AIJ) is a semi-positive definite matrix, then-a = (-aiji) is a semi-positive definite matrix.
By inference 1 Yes:
det (- A)≤
n
k= 1
∏(- akk)
$(- 1)ndetA≤(- 1)n
n
k= 1
* akk
$
det A≤
n
k= 1
∏akk, n is an odd number,
det A≥
n
k= 1
∏akk, n is an even number.
%
'
& amp
'
(
,
Complete the certificate.
References:
[1] Department of Mathematical Mechanics, Peking University. Advanced Algebra (3rd Edition) [M]. Beijing:
Higher Education Press, 2003.