yes, it is both symmetric as well as skew symmetric
In linear algebra, a skew-symmetric matrix is a square matrix .....'A'
I could be wrong but I do not believe that it is possible other than for the null matrix.
My knowledge limits to square matrices. The answer is yes, because 0 = -0
In a skew symmetric matrix of nxn we have n(n-1)/2 arbitrary elements. Number of arbitrary element is equal to the dimension. For proof, use the standard basis.Thus, the answer is 3x2/2=3 .
The normal of a square matrix refers to a matrix that commutes with its conjugate transpose, meaning that for a square matrix ( A ), it is considered normal if ( A A^* = A^* A ), where ( A^* ) is the conjugate transpose of ( A ). Normal matrices include categories such as Hermitian, unitary, and skew-Hermitian matrices. These matrices have important properties, such as having a complete set of orthonormal eigenvectors and being diagonalizable via a unitary transformation.
A skew symmetric matrix is a square matrix which satisfy, Aij=-Aji or A=-At
Let A be a matrix which is both symmetric and skew symmetric. so AT=A and AT= -A so A =- A that implies 2A =zero matrix that implies A is a zero matrix
In linear algebra, a skew-symmetric matrix is a square matrix .....'A'
I could be wrong but I do not believe that it is possible other than for the null matrix.
My knowledge limits to square matrices. The answer is yes, because 0 = -0
In a skew symmetric matrix of nxn we have n(n-1)/2 arbitrary elements. Number of arbitrary element is equal to the dimension. For proof, use the standard basis.Thus, the answer is 3x2/2=3 .
yes
#include<iostream.h>
Skew-Hermitian matrix defined:If the conjugate transpose, A†, of a square matrix, A, is equal to its negative, -A, then A is a skew-Hermitian matrix.Notes:1. The main diagonal elements of a skew-Hermitian matrix must be purely imaginary, including zero.2. The cross elements of a skew-Hermitian matrix are complex numbers having equal imaginary part values, and equal-in-magnitude-but-opposite-in-sign real parts.
They can be either. If they are roots of a real polynomial then purely imaginary would be symmetric and only real roots can be skew symmetric.
No, there cannot be any.
In a symmetric binomial distribution, the probabilities of success and failure are equal, resulting in a symmetric shape of the distribution. In a skewed binomial distribution, the probabilities of success and failure are not equal, leading to an asymmetric shape where the distribution is stretched towards one side.