From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: definite semidefinite matrix
Date: 3 May 1999 15:46:57 -0400
Newsgroups: sci.math
In article <925467749.857148@marvin>,
Jian Wang wrote:
>Are there any rule to tell if a given matrix a positive definite or positive
>semi-definite matrix, rather then by their definition?
>p.s. the definition:
> let A be a matrix, x a vector, and Q=, if Q>0 for x<>0, and Q = 0,
>for x = 0, then A is positive definite;
> let A be a matrix, x a vector, and Q=, if Q>=0 for x<>0, and Q =
>0, for x = 0, then A is positive semidefinite.
>
>Thanks for your attention.
n-by-n positive definite:
The following are equivalent (here ^H denotes Hermitian transpose)
(1)
Over real scalars, assume symmetry of A; over complex scalars, Hermitian
symmetry will come out as a bonus. The defining condition is
for every x other than zero, > 0
(2) Hermitian symmetry assumed: all eigenvalues are positive
(3) there is a matrix B of rank n (can be rectangular) such that
A = B^H * B
(this is useful if you know that A was put together this way)
(4) there is an invertible matrix T and a diagonal matrix D with positive
diagonal entries such that A = T^H * D * T
(4a) there is a unitary matrix U and a diagonal matrix D with positive
diagonal entries such that A = U * D * U^H
(5) there is a (previously identified as such) positive definite matrix C
and a matrix T of rank n such that A = T^H * C * T
(useful when T is rectangular)
(6) the A = L * D * L^H decomposition is successful with L unit lower
triangular, and D diagonal with positive diagonal entries,
(7) the Cholesky decomposition A = C * C^H is successful with C invertible
lower triangular,
(8) (Hermitian symmetry assumed): the leading subdetermimants
det(A(1:k,1:k)) (first k rows from the first k columns) are positive
(Sylvester's test)
(9) A = B + C where B is positive definite and C is positive semidefinite
(B and C previously identified as such)
-- see remakr below --
(10) A = B + C where B and C are positive semidefinite, and their
nullspaces have only 0 in common
and many others.
There is a sufficient condition, as mentioned elsewhere: assuming
Hermitian symmetry, A be strictly diagonally dominant with positive
diagonal entries, perhaps after an appropriate (positive) column scaling.
This means that every diagonal entry is greater than the sum of absolute
values of the other entries in the same row.
For semidefinite matrices, adjustments must be made, and some conditions
will be true only one way.
Example: the eigenvalues are allowed to be non-negative, rather than
positive.
Or: A = B^H * B where we impose no restriction on the rank of B.
Topologically, positive semidefinite matrices are limits of positive
definite matrices. So:
A is positive semidefinite if and only if for every r>0, the matrix A+rI
is positive definite.
Hope it helps, ZVK(Slavek)
==============================================================================
From: israel@math.ubc.ca (Robert Israel)
Subject: Re: Please help with intuitive meaning/use of 'positive definite'
Date: 25 Aug 1999 20:35:52 GMT
Newsgroups: sci.math
In article <37C2BB8B.D1A39B0F@jpl.nasa.gov>,
Tim Rogstad writes:
> Can someone help me with this? When a matrix is 'positive definite'
> (or 'definite', 'strictly definite', etc.) what does that really mean as
> far as applications? When is this property used? I need an intuitive
> explanation.
Well, there are lots of possible applications. To choose the one that
students usually encounter first: if H is the Hessian matrix of a function f
of several variables, i.e. the matrix of second partial derivatives
H_{ij} = (del^2 f)/(del x_i del x_j)
then definiteness of H is related to the classification of critical points
of f. Thus if f has a critical point at which H is positive definite,
that critical point is a local minimum of f.
A rather different application:
Consider a system of equal point masses connected by linear springs. The
differential equation governing this system (according to Newton's Laws of
Motion) is of the form
x" = -A x
where x is the vector consisting of all coordinates of all the bodies,
" is the second derivative with respect to time, and A is a certain matrix.
This is symmetric as a consequence of Newton's laws (the force exerted by
i on j is equal and opposite to the force exerted by j on i). If A has
an eigenvector v for eigenvalue r, then x = exp(sqrt(-r) t) v is a
solution. We don't expect solutions that grow exponentially, so
there shouldn't be any negative eigenvalues. Thus A should be positive
semidefinite (it will have 0 as an eigenvalue if the whole system, or
part of it, is free to move in some direction). The eigenvectors of A
correspond to the different "normal modes" of vibration of the system.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
==============================================================================
From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: Please help with intuitive meaning/use of 'positive definite'
Date: 26 Aug 1999 16:34:54 -0400
Newsgroups: sci.math
In article <37C2BB8B.D1A39B0F@jpl.nasa.gov>,
Tim Rogstad wrote:
:Hi,
: Can someone help me with this? When a matrix is 'positive definite'
:(or 'definite', 'strictly definite', etc.) what does that really mean as
:far as applications? When is this property used? I need an intuitive
:explanation.
:
:Thanks,
:Tim
(Assumed: the matrix A has real entries)
It's the quadratic form defined by the matrix: a function sending a
vector x to a number Q(x) = x'*A*x (x' is the transpose) , that is closer
to applications.
The level sets (curves for 2-by-2 matrices, surfaces for 3-by-3 matrices,
then they are called hypersurfaces) od a positive (or negative) definite
form, if not empty or a single point, are ellipses/ellipsoids, in contrast
with other (mainly indefinite) froms when the level sets are unbounded.
This is a generic small-scale behavior of a smooth function (of many
variables) near a local minimum, and positive definiteness of the matrix
of the second derivatives (this matrix is called the Hessian) is a part of
the test for a local minimum.
In mechanics, the positive definiteness of the Hessian near a stationary
point of the potential indicates a stable equilibrium. And the kinetic
energy in generalized coordinates is a positive definite form in the
generalized velocities.
Hope this is applied enough.
Cheers, ZVK(Slavek).