Symmetric and orthogonal matrices in Rn n Hermitian and unitary matrices in Cn n Defn: if AT = A()A= symmetric Defn: if AH = A()A= Hermitian A= symmetric =)Ais a square matrix A= Hermitian =)Ais a square matrix a pure complex matrix cannot be Hermitian (the diagonal must have real entries) A= symmetric =) i 2R;8i A= Hermitian =) i 2R;8i

Jun 03, 2017 Pauli and Dirac matrices | Mathematics for Physics Δ It is important to remember that the Dirac matrices are matrix representations of an orthonormal basis of the underlying vector space used to generate a Clifford algebra. So the Dirac and chiral bases are different representations of the orthonormal basis which generates the matrix representation \({C\mathbb{^{C}}(4)\cong\mathbb{C}(4)}\) acting on vectors (spinors) in \({\mathbb{C}^{4}}\). Determine if matrix is Hermitian or skew-Hermitian

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Traceless Hermitian Matrices | Physics Forums Feb 04, 2017 Gauge fields -- why are they traceless hermitian A gauge field is introduced in the theory to preserve local gauge invariance. And this field (matrix) is expanded in terms of the generators, which is possible because the gauge field is traceless hermitian.

square matrix A is Hermitian if and only if the following two conditions are met. 1. The entries on the main diagonal of A are real. 2. The entry in the ith row and the jth column is the complex conjugate of the entry in the jth row and ith column. EXAMPLE 4 Hermitian Matrices

$\begingroup$ The diagonal matrix $\textrm{diag}(1+i,1+i,-1-i,-1-i)$ has determinant $-4$, and then in general build the matrix from such blocks. $\endgroup$ – Christian Remling Apr 3 '19 at 19:18 2 A hermitian matrix of unit determinant and having positive eigenvalues can be uniquely expressed as the exponential of a traceless hermitian matrix, and therefore the topology of this is that of (n2 − 1)-dimensional Euclidean space. A hermitian matrix of unit determinant and having positive eigenvalues can be uniquely expressed as the exponential of a traceless hermitian matrix, and therefore the topology of this is that of (n 2 − 1)-dimensional Euclidean space. Since SU(n) is simply connected, we conclude that SL(n, C) is also simply connected, for all n. The Pauli matrices also form a basis for the vector space of traceless hermitian \({2\times2}\) matrices, which means that \({i\sigma_{i}}\) is a basis for the vector space of traceless anti-hermitian matrices \({su(2)\cong so(3)}\).