Definition 6.2.1.
The matrix
is a positive definite matrix if
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for all nonzero vector
.
Example 6.2.1. The
matrix
![]()
is a positive definite one since for
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Problem 6.2.1.* Show that the matrix

is positive definite.
Proposition 6.2.1.
If
is a positive definite matrix and
the column vectors of the matrix
are linearly independent,
then the matrix
![]()
is also positive definite.
Proof. If for the vector
it holds the relation
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then
![]()
and from the positive definiteness
of the matrix A it follows that
Since the column vectors of the matrix X are linearly
independent, then from
it follows that
Hence from conditions
and
it follows that
i.e., the matrix B is positive definite.
Corollary 6.2.1. If
the matrix
is positive definite, then all
the submatrices of the matrix A obtained by deleting the rows and
columns of the matrix A with the same numbers are positive
definite and all the elements on the leading diagonal of the matrix
are positive.
Proof. If
is a vector with natural number coordinates satisfying the condition
![]()
then
![]()
is a matrix got from the unit matrix In by taking the
column-vectors with indices
Hence the column-vectors of the matrix X are linearly
independent, and proposition 6.2.1
implies that the matrix XTAX is positive
definite. The matrix XTAX is a submatrix of the matrix
A obtained from the rows and columns with numbers
of the matrix A. Therefore, all the submatrices of the matrix A
obtained by deleting the rows and columns of the matrix A with the
same numbers are positive definite. Taking k=1, we get the second
part of the statement.
Corollary 6.2.2.
If
is positive definite, then the matrix A has a decomposition A=LDMT
and all the leading diagonal elements of the matrix D are positive.
Proof. On the ground of corollary
6.2.1, all the submatrices
of the matrix A are positive definite, and, therefore, regular matrices,
and proposition 6.1.1 implies
the existence of the LDMT
decomposition. Taking X=L-T in proposition
6.2.1, we find that the matrix
![]()
is positive definite. Since the matrix MTL-T
is an upper triangular matrix with the unit diagonal, the matrices B
and D have the same leading diagonal and the elements on it must
be positive, provided that B is positive
definite.
Proposition 6.2.2
(Cholesky factorization). If the matrix
is symmetric and positive definite,
then there exists exactly one lower triangular matrix G with the
positive leading diagonal such that
![]()
Proof. In virtue of proposition
6.1.2, there exist and are uniquely defined the lower triangular matrix
L with the unit diagonal and the diagonal matrix
such that it holds the decomposition (3), i.e.,
A=LDLT. The corollary
6.2.2 provides that elements dk of the matrix D
are positive. Therefore, the matrix
![]()
is a lower triangular matrix with the positive leading diagonal, and equality
(5) holds. The uniqueness of the the factorization follows
from the uniqueness of the decomposition (3).
The factorization (5) is known as the
Cholesky factorization. The
matrix G is called the Cholesky triangular matrix of the
matrix A. To solve the system of equations
![]()
having the symmetric and positive definite
matrix A, one has to find the Cholesky triangular matrix of
the matrix A. Secondly, one has to solve the system with the triangular
matrix
![]()
Thirdly, one has to solve the system
![]()
The Cholesky factorization can be found step by step.
Proposition 6.2.3.
If the matrix
is symmetric and positive definite,
then, denoting
![]()
the matrix A can be expressed by
![]()
where
The matrix
is positive definite. If
![]()
then A=GGT, where
![]()
Proof. Let us check the accurancy of the decomposition
(6):
![]()
![]()
![]()
If
![]()
then
![]()
![]()
Since the matrix A is positive definite and the column-vector system
of the matrix X is linearly
independent, proposition 6.2.1 implies the positive definiteness of
the matrix
![]()
and from corollary 6.2.1 it follows that the matrix
is linewise positive definite. So we can, analogously to the partition
of the matrix A into blocks, decompose also the matrix
into blocks, etc.
Example 6.2.2. Let us find the LU,
LDMT, LDLT
and Cholesky factorizations of
the matrix

The principal minors of the matrix A are nonzero.
We find

and

and also

Knowing the LU factorization of
the matrix A, we will find the LDMT
decomposition, LDLT
decomposition and Cholesky factorization
of it:

and

Let us find the Cholesky factorization
of the matrix A also step by step using the algorithm given
in proposition 6.2.3. Since on the first
step
![]()
then
![]()
On the next step,
![]()
and
![]()
Therefore,
![]()
and
![]()
and

Problem 6.2.2.* Find the Cholesky
factorization of the positive definite
matrix

Problem 6.2.3.* Solve the system of
equations
where

when the Cholesky factorization of
the matrix A is given
