## Announce

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## Matrix

### Positive definite

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

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>, :$\textbf{z}^{*} M \textbf{z} > 0$. Note that the quantity $z^{*} M z$ is always real because $M$ is a Hermitian matrix.

All eigenvalue?s $\lambda_i$ of $M$ are positive.

## Eigenvalue

### Fundamental

:$A \mathbf{x} = \lambda \mathbf{x}$ :$(A - \lambda I) \mathbf{x} = 0$ :$\det(A - \lambda I) = 0 \,$

### Eigendecomposition

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

The eigenvectors $q_i \,\, (i = 1, ..., N)$ are usually normalized (orthonomal eigenvectors)

:$\mathbf{A}^{-1}=\mathbf{Q}\mathbf{\Lambda}^{-1}\mathbf{Q}^{-1}$

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