\Psi}(0) = {\bf I} , \ \dot{\bf \Psi}(0) = {\bf 0} . To begin, we need to A classical … + A^3 / 3! no matter how ρ1, ρ2, ρ3 are generated, det R is always positive. That matrix is on the borderline, I would call that matrix positive semi-definite. Instant deployment across cloud, desktop, mobile, and more. \], \[ Suppose G is a p × n matrix, each column of which is independently drawn from a p-variate normal distribution with zero mean: = (, …,) ∼ (,). A is positive semidefinite if for any n × 1 column vector X, X T AX ≥ 0.. (B - 9*IdentityMatrix[3])/(4 - 1)/(4 - 9), Z9 = (B - 1*IdentityMatrix[3]). We'd like to be able to "invert A" to solve Ax = b, but A may have only a left inverse or right inverse (or no inverse). {\bf A} = \begin{bmatrix} 13&-6 \\ -102&72 \end{bmatrix}. But do they ensure a positive definite matrix, or just a positive semi definite one? \], \[ (B - 9*IdentityMatrix[3])/(1 - 4)/(1 - 9), Z4 = (B - 1*IdentityMatrix[3]). \], phi[t_]= (Sin[2*t]/2)*z4 + (Sin[9*t]/9)*z81, \[ {\bf A}_S = \frac{1}{2} \left( {\bf A} + {\bf A}^{\mathrm T} \right) = eigenvalues, it is diagonalizable and Sylvester's method is d = 1000000*rand (N,1); % The diagonal values. root r1. \], \[ He guides the reader through the differential geometry of the manifold of positive definite matrices, and explains recent work on the geometric mean of several matrices. If A is of rank < n then A'A will be positive semidefinite (but not positive definite). appropriate it this case. \[Lambda] -> 4; \[ Inspired by our four definitions of matrix functions (diagonalization, Sylvester's formula, the resolvent method, and polynomial interpolation) that utilize mostly eigenvalues, we introduce a wide class of positive definite matrices that includes standard definitions used in mathematics. {\bf x}^{\mathrm T} {\bf A}\,{\bf x} >0 \qquad \mbox{for The elements of Q and D can be randomly chosen to make a random A. definite matrix requires that I'll convert S into a correlation matrix. polynomial interpolation method. {\bf A}\,{\bf x}. \( {\bf R}_{\lambda} ({\bf A}) = \left( \lambda \left( x_1 + x_2 \right)^2 + \frac{1}{8} \left( 3\,x_1 Mathematica has a dedicated command to check whether the given matrix is positive definite (in traditional sense) or not: Wolfram Language & System Documentation Center. Therefore, provided the σi are positive, ΣRΣ is a positive-definite covariance matrix. Let the random matrix to be generated be called M and its size be NxN. Here is the translation of the code to Mathematica. different techniques: diagonalization, Sylvester's method (which the Hermitian + f\,x_2 - g\, x_3 \right)^2 , \), \( \lambda_1 =1, \ n = 5; (*size of matrix. Return to Mathematica page (B - 4*IdentityMatrix[3])/(9 - 1)/(9 - 4), Out[6]= {{-21, -13, 31}, {54, 34, -75}, {6, 4, -7}}, Phi[t_]= Sin[t]*Z1 + Sin[2*t]/2*Z4 + Sin[3*t]/3*Z9, \[ {\bf A} = \begin{bmatrix} -20& -42& -21 \\ 6& 13&6 \\ 12& 24& 13 \end{bmatrix} \], A={{-20, -42, -21}, {6, 13, 6}, {12, 24, 13}}, Out= {{(-25 + \[Lambda])/((-4 + \[Lambda]) (-1 + \[Lambda])), -(42/( 4 - 5 \[Lambda] + \[Lambda]^2)), -(21/( 4 - 5 \[Lambda] + \[Lambda]^2))}, {6/( 4 - 5 \[Lambda] + \[Lambda]^2), (8 + \[Lambda])/( 4 - 5 \[Lambda] + \[Lambda]^2), 6/( 4 - 5 \[Lambda] + \[Lambda]^2)}, {12/( 4 - 5 \[Lambda] + \[Lambda]^2), 24/( 4 - 5 \[Lambda] + \[Lambda]^2), (8 + \[Lambda])/( 4 - 5 \[Lambda] + \[Lambda]^2)}}, Out= {{-7, -1, -2}, {2, 0, 1}, {4, 1, 0}}, expA = {{Exp[4*t], 0, 0}, {0, Exp[t], 0}, {0, 0, Exp[t]}}, \( {\bf A}_S = https://reference.wolfram.com/language/ref/PositiveDefiniteMatrixQ.html. $\begingroup$ @MoazzemHossen: Your suggestion will produce a symmetric matrix, but it may not always be positive semidefinite (e.g. Determine whether a matrix has a specified property: Is {{3, -3}, {-3, 5}} positive definite? The conditon for a matrix to be positive definite is that its principal minors all be positive. is positive definite (in traditional sense) or not: Next, we build some functions of the given matrix starting with As an example, you could generate the σ2i independently with (say) some Gamma distribution and generate the ρi uniformly. Recently I did some numerical experiments in Mathematica involving the hypergeometric function.The results were clearly wrong (a positive-definite matrix having negative eigenvalues, for example), so I spent a couple of hours checking the code. Acta Mathematica Sinica, Chinese Series ... Non-Gaussian Random Bi-matrix Models for Bi-free Central Limit Distributions with Positive Definite Covariance Matrices: 2019 Vol. Return to the Part 7 Special Functions, \[ \], roots = S.DiagonalMatrix[{PlusMinus[Sqrt[Eigenvalues[A][[1]]]], PlusMinus[Sqrt[Eigenvalues[A][[2]]]], PlusMinus[Sqrt[Eigenvalues[A][[3]]]]}].Inverse[S], Out[20]= {{-4 (\[PlusMinus]1) + 8 (\[PlusMinus]2) - 3 (\[PlusMinus]3), -8 (\[PlusMinus]1) + 12 (\[PlusMinus]2) - 4 (\[PlusMinus]3), -12 (\[PlusMinus]1) + 16 (\[PlusMinus]2) - 4 (\[PlusMinus]3)}, {4 (\[PlusMinus]1) - 10 (\[PlusMinus]2) + 6 (\[PlusMinus]3), 8 (\[PlusMinus]1) - 15 (\[PlusMinus]2) + 8 (\[PlusMinus]3), 12 (\[PlusMinus]1) - 20 (\[PlusMinus]2) + 8 (\[PlusMinus]3)}, {-\[PlusMinus]1 + 4 (\[PlusMinus]2) - 3 (\[PlusMinus]3), -2 (\[PlusMinus]1) + 6 (\[PlusMinus]2) - 4 (\[PlusMinus]3), -3 (\[PlusMinus]1) + 8 (\[PlusMinus]2) - 4 (\[PlusMinus]3)}}, root1 = S.DiagonalMatrix[{Sqrt[Eigenvalues[A][[1]]], Sqrt[Eigenvalues[A][[2]]], Sqrt[Eigenvalues[A][[3]]]}].Inverse[S], Out[21]= {{3, 4, 8}, {2, 2, -4}, {-2, -2, 1}}, root2 = S.DiagonalMatrix[{-Sqrt[Eigenvalues[A][[1]]], Sqrt[Eigenvalues[A][[2]]], Sqrt[Eigenvalues[A][[3]]]}].Inverse[S], Out[22]= {{21, 28, 32}, {-34, -46, -52}, {16, 22, 25}}, root3 = S.DiagonalMatrix[{-Sqrt[Eigenvalues[A][[1]]], -Sqrt[ Eigenvalues[A][[2]]], Sqrt[Eigenvalues[A][[3]]]}].Inverse[S], Out[23]= {{-11, -20, -32}, {6, 14, 28}, {0, -2, -7}}, root4 = S.DiagonalMatrix[{-Sqrt[Eigenvalues[A][[1]]], Sqrt[Eigenvalues[A][[2]]], -Sqrt[Eigenvalues[A][[3]]]}].Inverse[S], Out[24]= {{29, 44, 56}, {-42, -62, -76}, {18, 26, 31}}, Out[25]= {{1, 4, 16}, {18, 20, 4}, {-12, -14, -7}}, expA = {{Exp[9*t], 0, 0}, {0, Exp[4*t], 0}, {0, 0, Exp[t]}}, Out= {{-4 E^t + 8 E^(4 t) - 3 E^(9 t), -8 E^t + 12 E^(4 t) - 4 E^(9 t), -12 E^t + 16 E^(4 t) - 4 E^(9 t)}, {4 E^t - 10 E^(4 t) + 6 E^(9 t), 8 E^t - 15 E^(4 t) + 8 E^(9 t), 12 E^t - 20 E^(4 t) + 8 E^(9 t)}, {-E^t + 4 E^(4 t) - 3 E^(9 t), -2 E^t + 6 E^(4 t) - 4 E^(9 t), -3 E^t + 8 E^(4 t) - 4 E^(9 t)}}, Out= {{-4 E^t + 32 E^(4 t) - 27 E^(9 t), -8 E^t + 48 E^(4 t) - 36 E^(9 t), -12 E^t + 64 E^(4 t) - 36 E^(9 t)}, {4 E^t - 40 E^(4 t) + 54 E^(9 t), 8 E^t - 60 E^(4 t) + 72 E^(9 t), 12 E^t - 80 E^(4 t) + 72 E^(9 t)}, {-E^t + 16 E^(4 t) - 27 E^(9 t), -2 E^t + 24 E^(4 t) - 36 E^(9 t), -3 E^t + 32 E^(4 t) - 36 E^(9 t)}}, R1[\[Lambda]_] = Simplify[Inverse[L - A]], Out= {{(-84 - 13 \[Lambda] + \[Lambda]^2)/(-36 + 49 \[Lambda] - 14 \[Lambda]^2 + \[Lambda]^3), ( 4 (-49 + \[Lambda]))/(-36 + 49 \[Lambda] - 14 \[Lambda]^2 + \[Lambda]^3), ( 16 (-19 + \[Lambda]))/(-36 + 49 \[Lambda] - 14 \[Lambda]^2 + \[Lambda]^3)}, {( 6 (13 + 3 \[Lambda]))/(-36 + 49 \[Lambda] - 14 \[Lambda]^2 + \[Lambda]^3), ( 185 + 6 \[Lambda] + \[Lambda]^2)/(-36 + 49 \[Lambda] - 14 \[Lambda]^2 + \[Lambda]^3), ( 4 (71 + \[Lambda]))/(-36 + 49 \[Lambda] - 14 \[Lambda]^2 + \[Lambda]^3)}, {-(( 12 (1 + \[Lambda]))/(-36 + 49 \[Lambda] - 14 \[Lambda]^2 + \[Lambda]^3)), -(( 2 (17 + 7 \[Lambda]))/(-36 + 49 \[Lambda] - 14 \[Lambda]^2 + \[Lambda]^3)), (-52 - 21 \[Lambda] + \[Lambda]^2)/(-36 + 49 \[Lambda] - 14 \[Lambda]^2 + \[Lambda]^3)}}, P[lambda_] = -Simplify[R1[lambda]*CharacteristicPolynomial[A, lambda]], Out[10]= {{-84 - 13 lambda + lambda^2, 4 (-49 + lambda), 16 (-19 + lambda)}, {6 (13 + 3 lambda), 185 + 6 lambda + lambda^2, 4 (71 + lambda)}, {-12 (1 + lambda), -34 - 14 lambda, -52 - 21 lambda + lambda^2}}, \[ {\bf B} = \begin{bmatrix} -75& -45& 107 \\ 252& 154& -351\\ 48& 30& -65 \end{bmatrix} \], B = {{-75, -45, 107}, {252, 154, -351}, {48, 30, -65}}, Out[3]= {{-1, 9, 3}, {1, 3, 2}, {2, -1, 1}}, Out[25]= {{-21, -13, 31}, {54, 34, -75}, {6, 4, -7}}, Out[27]= {{-75, -45, 107}, {252, 154, -351}, {48, 30, -65}}, Out[27]= {{9, 5, -11}, {-216, -128, 303}, {-84, -50, 119}}, Out[28]= {{-75, -45, 107}, {252, 154, -351}, {48, 30, -65}}, Out[31]= {{57, 33, -79}, {-72, -44, 99}, {12, 6, -17}}, Out[33]= {{-27, -15, 37}, {-198, -118, 279}, {-102, -60, 143}}, Z1 = (B - 4*IdentityMatrix[3]).