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Positive definite

Let M be an <math>n \times n</math> Hermitian matrix?. Denote the transpose of a vector <math>a</math> by <math>a^{T}</math>, and the conjugate transpose? by <math>a^{*}</math>.

The matrix M is positive definite if and only if? it satisfies any of the following equivalent properties:

For all non-zero complex vectors zC<sup>n</sup>, :<math>\textbf{z}^{*} M \textbf{z} > 0</math>. Note that the quantity <math>z^{*} M z</math> is always real because <math>M</math> is a Hermitian matrix.

All eigenvalue?s <math>\lambda_i</math> of <math>M</math> are positive.



:<math>A \mathbf{x} = \lambda \mathbf{x}</math> :<math>(A - \lambda I) \mathbf{x} = 0</math> :<math>\det(A - \lambda I) = 0 \,</math>


Let A be a square (N×N) matrix with N linearly independent? eigenvectors, <math>q_i \,\, (i = 1, ..., N).</math> Then A can be matrix decomposition|factorized? as :<math>\mathbf{A}=\mathbf{Q}\mathbf{\Lambda}\mathbf{Q}^{-1} </math> where Q is the square (N×N) matrix whose i<sup>th</sup> column is the eigenvector <math>q_i</math> of A and Λ is the diagonal matrix? whose diagonal elements are the corresponding eigenvalues, i.e., <math>\Lambda_{ii}=\lambda_i</math>.

The eigenvectors <math>q_i \,\, (i = 1, ..., N)</math> are usually normalized (orthonomal eigenvectors)

:<math>\mathbf{A}^{-1}=\mathbf{Q}\mathbf{\Lambda}^{-1}\mathbf{Q}^{-1} </math>

Because Λ is a diagonal matrix?, its inverse is easy to calculate: :<math>\left[\Lambda^{-1}\right]_{ii}=\frac{1}{\lambda_i}</math>