When is matrix positive definite

Schaum's Outline of Theory and Problems of Matrices. New York: Schaum, p. Golub, G. Gradshteyn, I. Tables of Integrals, Series, and Products, 6th ed. Johnson, C. Monthly 77 , Lindell, I. Methods for Electromagnetic Field Analysis. New York: Clarendon Press, Marcus, M. Introduction to Linear Algebra. New York: Dover, p. Pease, M. Methods of Matrix Algebra. Here is another way to see this. This gives some intuition for why "most" positive definite matrices are not diagonally dominant.

This is in part a rehash of Noah Stein answer, without too much clutter. More importantly, it gives you an algorithmic way to determine what type of matrix you have. It's easy if you follow Cholesky method in here. A squared matrix is positive definite if it is symmetric!

If this is true, then see the reference! Nice and neat! For the two statements mentioned above, the first one is true just based on the Gershgorin Circle Theorem. But actually, the second statement is true if you add one more condition on A: matrix A is irreducible. Then the counter-example provided below is no sense any more since they are reducible.

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Moreover, can be chosen to be real since a real solution to the equation is guaranteed to exist because is rank-deficient by the definition of eigenvalue. Then, we have where is the norm of. Since is an eigenvector,. Moreover, by the definiteness property of the norm,. Thus, we have because by the hypothesis that is positive definite we have demonstrated above that the quadratic form involves a real vector , which is required in our definition of positive definiteness.

We have proved that any eigenvalue of is strictly positive, as desired. Let us now prove the "if" part, starting from the hypothesis that all the eigenvalues of are strictly positive real numbers. Since is real and symmetric, it can be diagonalized as follows: where is orthogonal and is a diagonal matrix having the eigenvalues of on the main diagonal as proved in the lecture on normal matrices.

The eigenvalues are strictly positive, so we can write where is a diagonal matrix such that its -th entry satisfies for. Therefore, and, for any vector , we have The matrix , being orthogonal, is invertible hence full-rank.

The matrix is diagonal hence triangular and its diagonal entries are strictly positive, which implies that is invertible hence full-rank by the properties of triangular matrices. The product of two full-rank matrices is full-rank. Therefore, is full-rank. Thus, because. By the positive definiteness of the norm, this implies that and, as a consequence, Thus, is positive definite. In what follows positive real number means a real number that is greater than or equal to zero.

Proposition A real symmetric matrix is positive semi-definite if and only if all its eigenvalues are positive real numbers. We do not repeat all the details of the proof and we just highlight where the previous proof for the positive definite case needs to be changed. The first change is in the "only if" part, where we now have because by the hypothesis that is positive semi-definite.

The second change is in the "if part", where we have because the entries of are no longer guaranteed to be strictly positive and, as a consequence, is not guaranteed to be full-rank. It follows that.