Questions tagged [positive-definite]

For questions about positive definite real or complex matrices. For questions about positive semi-definite matrices, use the (positive-semidefinite) tag.

This tag is for questions about positive definite matrices with real or complex entries. A square matrix $M \in \mathbf{F}^{n \times n}$ ($\mathbf{F} = \mathbf{R}$ or $\mathbf{C}$) is positive definite if $$ \text{for all } x \in \mathbf{F}^n \setminus \{0\}, x^\dagger M x > 0. \tag{1}$$ Here $x^\dagger$ denotes the transpose if $x$ is real and the conjugate-transpose if $x$ is complex.

If we replace $(1)$ with $$ \text{for all } x \in \mathbf{F}^n, x^\dagger M x \ge 0 $$ then $M$ is said to be positive semi-definite. All positive definite matrices are positive semi-definite. Questions about positive semi-definite matrices not specifically about positive definite matrices should use the tag instead or in conjunction.

If $\mathbf{F} = \mathbf{C}$ then $M$ is positive definite if and only if $M^\dagger = M$ and every eigenvalue of $M$ is a positive real number. If $\mathbf{F} = \mathbf{R}$ then it is not necessary that $M^\dagger = M$, for instance $$ M = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} $$ is a positive definite real matrix but not symmetric. Some authors require that a positive definite matrix be symmetric.

Some authors use a weaker form of $(1)$, namely $$ \text{for all } x \in \mathbf{F}^n \setminus \{0\}, \operatorname{Re}(x^\dagger M x) > 0. $$ With this definition it is no longer necessary that $M^\dagger = M$, even if $\mathbf{F} = \mathbf{C}$.

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Why the derivative is negative definite?

Why the derivative is negative definite? Shouldn't be negative semi definite? It only depends on the $x$.
CroCo
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What is the meaning of a nonnegative definite function?

I know non-negative definiteness for a matrix, but I am reading some notes that use it for a function $f \colon \mathbb{Z} \rightarrow \mathbb{R}$. I am not sure since it doesnt' say so explictily, but I inferred that the definition is something…
jansai
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A question about positive definite and symmetric matrices

Suppose $A$, $B$ are real $n \times n$ matrices, and $A$, $AB$ are symmetric with $A$ positive definite. Is the matrix $B$ diagonalizable and having real eigenvalues ? Could someone give me a hint about this question ?
user405606
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Does adding non-negative values to diagonal of positive definite matrix preserves positive definiteness?

Does adding non-negative values to the diagonal of a positive definite matrix preserves its positive definiteness? For example, $A$ is a symmetric positive definite matrix, and $D$ is a diagonal matrix with non-negative elements. Is $A+D$ positive…
Tan
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proof that covariance matrix of multivariate normal distribution is positive definite

I want to know the proof that covariance matrix of multivariate normal distribution is positive definite in order to have a pdf. To have a pdf, if x is a random vector of size n, 1) f(x) >=0 2) $$\int_{-\infty}^{\infty} f(x) dx$$ = 1 how can this…
Damelim
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Expressing a positive semi-definite matrix as sum of a diagonal and a rank one projector

Let $A\geq0$ be an $n\times n$ positive semi-definite matrix over $\mathbb{C}$, i.e., $x^* Ax\geq 0$ for all $x\in\mathbb{C}^n$. Given $A$, is it always possible to find real numbers $\alpha,\beta\geq0$, a diagonal matrix $D\geq 0$, and a vector…
Poor.Jack
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Given $A \in \mathbb{R}^{n \times n}$ and $B \in \mathbb{R}^{n \times p}$ with $rank(B)=p$ and $C(B)\subset C(A)$. Prove $B'AB$ is positive definite?

I want to show: Given $A \in \mathbb{R}^{n \times n}$ and $B \in \mathbb{R}^{n \times p}$ with $rank(B)=p$ and $C(B)\subset C(A)$. Prove $B'AB$ is positive definite, where $C(B)$ denotes the column space of matrix $B$. The steps below are what I…
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Which condition should A satisfy - Cholesky factorization

I have been given some matrix A that have size nxn and is nonsingular. I have to find out which condition A should satisfy such that it has a Cholesky factorization A = LL^T. Can someone help me with this? Because should A not just be positive…
DiscoCarl
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Find a positive definite non-convex function which is the difference of two positive definite functions

I'd like to find a pair of functions p, f such that $p(x) = 0.5 x^2 - 0.25 f(x)^2$, where $p$ is positive definite and non-convex and $f(0) = 0$. Is there any example that satisfies the above equation?
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Must a positive definite function be continuous?

Let $\Phi:\Omega\times\Omega\to\Bbb R$, with $\Omega\subseteq \Bbb R^d$. We say that $\Phi$ is positive definite if for every finite set $\{ x_1, x_2,\dots, x_n\}\subset\Omega$ of distinct points the $n\times n$ matrix $A$ whose entries $A_{ij}$…
User
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If $X^T$ is full rank, then $X^TX$ is positive definite?

I read that If $X^T$ is full column rank ($X$ is not necessarily square), then $X^TX$ is positive definite, and I'm trying to see why this is the case. I know that if $X^T$ is full column rank, then $A=X^TX$ is also full rank. We also know that $A$…
24n8
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Maximum of a positive definitive function

Let $f$ be a positive definitive function from $ \mathbb{R}^d \to \mathbb{R}$. in https://en.wikipedia.org/wiki/Positive-definite_function it is stated that $ |f(x)| \leq f(0) $. I couldnt find any proof. Can you help me?
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Positive definite in mean square sense

$x$ is a random vector variable and matrix $P$ is symmetric and deterministic. If $\mathbb{E}\{x^T P x\}>0$ for any non-zero $x$, can we say $P$ is positive definite? If it is not, can you give me a counterexample? $\mathbb{E}$ is the expectation…
wayne
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If $A$ is positive definite and $Ax\le b$ can I say that $x\le A^{-1}b$?

If I have a vector enequiality $Ax\le b$ where $x,b\in\mathbb{R}^n$ and $A$ is a $n\times n$ positive definite matrix, can I say that the following vector inequality is true? $$x\le A^{-1}b$$
AlephZero
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