# Determine if a matrix if positive definite

• onako
In summary, the given matrix A is a symmetric matrix with strictly positive entries and the diagonal entries are the sum of off-diagonal entries in the corresponding row/column. The proof for A being positive definite by showing x^TAx>0 for some non-zero vector x can be done by using the Gershgorin circle theorem, which shows that there are no negative eigenvalues but does not exclude the possibility of zero eigenvalues. However, the matrix 1 1 \ 1 1 is a counterexample as it is singular and not positive definite. The author of the book may have misinterpreted the positive definiteness of this type of matrix. It can be proven that the matrix is positive definite for n>3 due to the
onako
Given a symmetric matrix
$$A=\left(\begin{array}{ccccc} \sum a_{1s} & & & & \\ & \ddots & & a_{ij} \\ & & \ddots & & \\ &a_{ij} & & \ddots & \\ & & & & \sum w_{as} \end{array}\right) \in\mathbb{R}^{n\times n},$$
with strictly positive entries a_{ij}, and with the diagonal entries being sum of off-diagonal entries residing
in the corresponding row/column, how to proceed with the proof for A being positive definite,
$$x^TAx>0$$
for some non-zero vector x.

The http://en.wikipedia.org/wiki/Gershgorin_circle_theorem shows there are no negative eigenvalues, but it doesn't exclude the possibiltiy of zero eigenvalues (i.e. a singular matrix).

In fact the matrix
$$\begin{pmatrix}1 & 1 \\ 1 & 1 \end{pmatrix}$$
is singular, and therefore not positive definite.

Thanks for providing the example.
I guess the author of the book stating the above positive-definiteness on the given matrix type
somehow misinterpreted it.

onako said:
Thanks for providing the example.
I guess the author of the book stating the above positive-definiteness on the given matrix type
somehow misinterpreted it.
Actually, unless I'm mistaken, the matrix is invertible (hence positive definite) if n>3.

This is true because the matrix is diagonally dominant. There is a theorem that says a Hermitian diagonally dominant matrix with real nonnegative diagonal entries is positive semidefinite. A proof of this is found here http://planetmath.org/?op=getobj&from=objects&id=7483

It can be shown that the inequality holds for n>3, but not in general case, as is observed above.

## 1. What is a positive definite matrix?

A positive definite matrix is a square matrix where all the eigenvalues are positive. In other words, it is a matrix that satisfies a specific set of mathematical conditions that make it useful in a variety of applications, such as optimization and statistics.

## 2. How do you determine if a matrix is positive definite?

To determine if a matrix is positive definite, you can use the Cholesky decomposition method. This involves decomposing the matrix into a lower triangular matrix and its transpose, and then checking if the diagonal elements of the lower triangular matrix are all positive. If they are, then the original matrix is positive definite.

## 3. What is the significance of a positive definite matrix?

A positive definite matrix has many important properties that make it useful in various fields of mathematics and science. For example, it is used in optimization problems to find the minimum value of a function, and in statistics to generate multivariate random numbers.

## 4. Can a matrix be both positive definite and positive semidefinite?

No, a matrix cannot be both positive definite and positive semidefinite. A positive definite matrix has all positive eigenvalues, while a positive semidefinite matrix has all non-negative eigenvalues. Therefore, a matrix can only be one or the other, but not both.

## 5. Are there any other methods for determining if a matrix is positive definite?

Yes, there are other methods for determining if a matrix is positive definite, such as using the Sylvester's criterion or the principal minors test. However, the Cholesky decomposition method is the most commonly used and efficient method for determining positive definiteness.

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