Please consult the thread generation about a necessary condition for positive definite matrices I do

Thank you! To sum up, I still think that the second method given by Wangyanstorm is relatively simple and straightforward! Diagonal elements can be taken out directly.

Also, I have some questions about other methods. His method one: "All eigenvalues of positive definite matrices are greater than zero, and the traces of the matrix (i.e.,

Sum of main diagonal elements) = sum of all eigenvalues》0

I don't know if I don't understand that the sum of the main diagonal elements is 0, and then how can it be said that every element is greater than zero? His method three: "Give another method First, when n=1, it is obviously true.

Suppose that n=k is true. then, when n=k+1. then consider one of its nth order master and sub-formula, which is also positively definite.

The sum of its diagonal elements is greater than 0. If we look at another nth-order master, the elements of its diagonal elements are all greater than zero. In summary, all the elements of the diagonal element are greater than 0.

To sum up, the proposition is proven.

I probably didn't understand this method of induction, but I had a few questions1

Is the matrix corresponding to each nth order principal and sub-formula of a positively definite matrix necessarily positively definite? Is it also necessary to give the necessary explanations? 2

n=k. then, when n=k+1. then consider one of its nth order master and sub-formula, which is also positively definite.

The sum of its diagonal elements is greater than 0. Then examine another nth-order master-sub-formula"Is there a mistake here, what you said is that after considering an n-order master-sub-formula, after considering another n-order master-sub-formula, what I want to ask is, for a positive definite matrix, is there only one n-order master-subordinate? The method given by ken880602:

The sufficient and necessary condition for a positive definite matrix is that all the eigenvalues are positive, then the necessary condition for all eigenvalues to be positive is that the sum of the elements on the main diagonal is greater than 0, and the product is greater than 0. And then recursion: "I don't quite understand the recursion here, but what is the recursion referring to?" I'm sorry.,The line generation is more water.,Maybe some of the questions are very naïve.,Forgive me.,Hehe.

It is a necessary condition Landlord please note that it is emphasized that it is a necessary condition! The sufficient and necessary condition for a positive definite matrix is that all the eigenvalues are positive, then the necessary condition for all eigenvalues to be positive is that the sum of the elements on the main diagonal is greater than 0, and the product is greater than 0. Then retract the landlord to think about it

Positive definite matrix. All eigenvalues are greater than zero, and the traces of the matrix (i.e., the sum of the main diagonal elements) = the sum of all eigenvalues》0 View original post

First, it is proved that the inverse symmetric matrix of matrix a:

Because matrix a is positively definite, matrix a is symmetrical, i.e., a t=a;

And because (a)t=(a t).

So (a) t=a ; Therefore, the matrix a is inverse and is a symmetric matrix.

Then, it is proved that the inverse of matrix a is a positive definite matrix:

Since the matrix a is positively definite, then there is x that belongs to r, and x is not equal to 0, such that x tax>0;

For x ta x=x ta aa x=x t(a )t aa x=(a x) ta(a x), and a x is not equal to 0;

Therefore (a x) ta(a x) > 0, so x t a a x>0, then a is a positive definite matrix.

a is a quadratic matrix, then |a|>> 0, x 2+y 2, 1 eigenpolynomial |λe-a|= find the eigenvalue - x 0, y 0, x 2+y 2-1 0, then x 0, y 0, x 2+y 2, 1

The sufficient and necessary condition for a matrix to be positively definite is that all the sequential principal and sub-formulas of this matrix are greater than zero, but is there a bit of a problem with your problem Decision theorem 1: The sufficient and necessary condition for the symmetric matrix a to be positively definite is: all the eigenvalues of a are positive.

Decision theorem 2: A sufficient and necessary condition for the symmetric matrix a to be positive is that the principal and sub-formulas of each order of a are positive.

Decision Theorem 3: A is a sufficient and necessary condition for any matrix A to be positive and definite: A contracts with the unit matrix.

There are many conclusions, and the better one is that the sequential master and sub formulas are greater than zero.

You have to understand what a positive definite matrix is. Sufficient and necessary conditions for positive definite matrices: Decision theorem 1: A sufficient and necessary condition for the symmetric matrix a to be positive is that the eigenvalues of a are all positive.

Decision theorem 2: A sufficient and necessary condition for the symmetric matrix a to be positive is that the principal and sub-formulas of each order of a are positive decision theorem 3: A is a sufficient and necessary condition for any matrix a to be positive definite: a contract is in the unit matrix.

Properties of positive definite matrices:

1.The positive definite matrix must be non-singular. Definition of nonsingular matrices: If the determinant of the nth order matrix a is not zero, i.e., |a|≠0。

2.Any of the principal and submatrices of a positive definite matrix is also a positive definite matrix.

A symmetry matrix is not necessarily a positive definite matrix, such as a zero-element square.

There are no transformations, the following two formulas are two simple real numbers (transposed 1x1 matrices), of course equal, don't think of them as matrices anymore.

Properties of positive definite matrices: Let m be a symmetric matrix of real coefficients of order n, and if you quarrel with any ruler, why not zero vector x=(x 1,..x n), both have xmx 0, which is called m's positive definite

Because a is positively definite, therefore, for any hedge non-zero scattering line x=(x 1,..x_n) ,xax′0.Let x x = k, obviously k0 (x x every element is squared) then xaax = (xax) (xax) k0 then a 2 is a positive definite matrix, 9,

First of all, copy a theorem:

A positive definite <=> there is a reversible moment bai array c, such that the transpose du of a=c*c next proves your question:

Because a Zhengding dao

So there is an invertible matrix c such that a=c*c is transposed.

Let the inverse transpose of c = d

then d is reversible, and.

The inverse of a = the transposition of d*d (which is obtained by taking the inverse on both sides of the above equation), so the inverse of a is also positively definite.

And the accompaniment of a*a =|a|*e

So the accompaniment of a =|a|*A of the inverse.

Where|a|is the determinant of a and is a positive number.

That is, a positive number multiplied by a positive definite matrix, so it is positively definite.

kdlx2006 | 2008-09-0590